# NCERT Solutions for Class 9 Mathematics Chapter 8 -Quadrilaterals

Mathematics is becoming more and more significant  due to its various  applications in our daily lives.. It has been applied in many fields of sciences and technology, which has accelerated  its growth in  various applications. As a result, professionals suggest  focusing on Mathematics  helps us to improvise our reasoning, analytical skills  and  problem solving skills.  It is turning out to be a crucial subject in the primary and secondary levels. Hence, it has become part of many competitive examinations.

The main topics covered in this chapter are the basics of the parallelogram and its properties, the key concepts of rhombus and all the properties associated, the applications of the rectangle as well as square and their properties and so on..

NCERT Solutions Class 9 Mathematics Chapter 8  is based on the latest CBSE syllabus.   They are effective in clearing concepts and provide authentic, reliable and to the point answers. Students are advised to follow NCERT books first and then refer to NCERT Solutions.  The solutions include everything you might be searching for during last minute preparation. It saves time to a great extent. . Hence, many teachers advise students to follow NCERT Solutions for Class 9 Mathematics Chapter 8 provided by Extramarks.

Extramarks’ website is one of the fastest growing online platforms for all primary school and secondary school studies. It is trusted by lakhs of teachers and students through its relentless services. You can find all the study material right from learning to developing skills to revising on the Extramarks website. It’s a one stop solution to all your problems.

## Key Topics Covered In NCERT Solutions for Class 9 Mathematics Chapter 8

An enclosed figure with four sides is called quadrilaterals. You have learnt about the different types of quadrilaterals in your lower grades. Students are familiar with basic geometrical shapes such as  square, rhombus, rectangle and parallelogram. By now, you can easily differentiate between different quadrilaterals by just looking at them. But have you ever wondered how is it possible for you to do so?

The answer lies in the shape and size of the quadrilaterals.. It’s because you know that there are certain properties governing each quadrilateral which differentiate one from the other. For example, you already know from the previous classes that ‘A square has all the four sides equal’.

In Advanced Mathematics, you need to  learn about  its properties  and find proof of the quadrilaterals.  You will find everything in detail and get more practice  regarding this topic by following  the NCERT Solutions for Class 9 Mathematics Chapter 8 available on the Extramarks’ website.

This chapter will upgrade students’ logical and analytical skills. The way the solutions are written will help you to grasp the topics completely and will lay a strong foundation for Mathematics in  higher classes.Hence, they will learn to solve any problem in a smart way. Students are advised to make full use of resources to make the most of it.

In a way Extramarks promotes learning by encouraging the students to be great learners and try to feed their insatiable curiosity through NCERT Solutions

### Introduction

In the previous chapter, we learnt about the triangle and its properties. In the chapter, we will cover quadrilaterals,  types of quadrilaterals and their properties.

When four non-collinear points make a closed figure, then it is called a Quadrilateral. It contains four sides, four angles and four vertices.

### Angle sum property of a Quadrilateral

In this section, we will learn about the angles of the Quadrilateral. You can recall from the previous classes that the sum of all angles of a Quadrilateral is 360 degrees.

To take your learning beyond books and understand better , we recommend NCERT Solutions for Class 9 Mathematics Chapter 8 available on the Extramarks’ website.

In this section, we will cover the types of the quadrilaterals covered in this chapter

• Trapezium

When one pair of the opposite side of the quadrilateral is parallel, it is called a trapezium.

• Parallelogram

When both pairs of the opposite sides of the quadrilateral are parallel, it is called a parallelogram.

Types of parallelograms like

• Rectangular.

When one of the angles of a parallelogram is the right angle, it is called a  rectangle.

• Rhombus

When all sides are equal in the parallelogram, it is called a  rhombus.

• Square

When one of the angles is a right angle, and all sides are equal in the parallelogram, it is called a square.

• Kite

When two pairs of adjacent sides are equal in a quadrilateral, it is called a  kite. It is not a parallelogram.

For more details, please visit our Extramarks’ website and refer to NCERT Solutions for Class 9 Mathematics Chapter 8..

Properties of a Parallelogram

An essential prerequisite to learning the  properties of the parallelogram , it is necessary to brush up these  theorems:

• Theorem1: A diagonal is a parallelogram that divides it into two congruent triangles.
• Theorem 2: In a parallelogram, opposite sides are equal.
• Theorem 3: If each pair of the opposite side of the quadrilateral is equal, then it is a parallelogram.
• Theorem 4: In a parallelogram, opposite angles are equal.
• Theorem 5: If in a quadrilateral, each pair of opposite angles is equal, then it is a parallelogram.
• Theorem 6: The diagonals of a parallelogram bisect each other.
• Theorem 7: If the diagonals of a quadrilateral bisect each other, then it is a parallelogram.

Another condition for a Quadrilateral to be a Parallelogram

We have already  studied  the properties of a  parallelogram.  A quadrilateral must meet certain criteria to be a parallelogram. They can be best understood with the help of the following theorem:

• Theorem: A quadrilateral is a parallelogram if a pair of opposite sides is equal and parallel.

The  Mid-point Theorem

So far, you have learnt about the triangles and the quadrilaterals. Now we will turn to  the midpoint of the sides of the triangle. You can  understand these concepts better  with the help of the theorems given  below-

• Theorem1: The line segment joining the mid- point of the two sides of a triangle is parallel to the third side.
• Theorem 2: The line drawn through the midpoint point of one side of a triangle parallel to another side bisects the third side.

Summary

In this chapter, we looked at  the quadrilaterals, their definitions, the sum of the angles of the quadrilateral,   its properties  and  key theorems required to understand  geometry.  We can sum up the following concepts:

1. Quadrilateral:  When four non-collinear points make a closed figure, then it is called a quadrilateral. It contains four sides, four angles and four vertices.
2.  The sum of all the angles of the quadrilateral is 360 degrees.
1.  Trapezium
2. Parallelogram
1. Rectangle
2. Rhombus
3. Square
3. Kite
• The properties of a  parallelogram

This chapter has seven important theorems.  .

•  The conditions for a quadrilateral to be a parallelogram
•  The mid-point theorem

NCERT Solutions for Class 9 Mathematics Chapter 8: Exercise &  Solutions

It requires a lot of practice to be a Mathematics wizard to solve advanced levels of questions correctly. The exercises given in the NCERT textbook have different types  of questions to check your understanding. You need endless practice and patience   to be strong in this subject.

The complete Class 9 Mathematics Chapter 8 exercises and solutions have been included in the NCERT Solutions for Class 9 Mathematics Chapter 8, available on the Extramarks website. It has a brief overview of all the exercises along with the solutions designed as per the  CBSE curriculum and guidelines.

Extramarks which tries to do away with rote learning and supplements their studies  with experiential learning and other innovative educational materials. Click on the  links below have to exercise specific questions and solutions for NCERT Solutions for Class 9 Mathematics Chapter 8:

• Chapter 8: Exercise 8.1 Question and answers
• Chapter 8: Exercise 8.2 Question and answers

Along with NCERT Solutions for Class 9 Mathematics Chapter 8, students can explore NCERT Solutions on our Extramarks’ website for all primary and secondary classes.

• NCERT Solutions Class 1
• NCERT Solutions Class 2
• NCERT Solutions Class 3
• NCERT Solutions Class 4
• NCERT Solutions Class 5
• NCERT Solutions Class 6
• NCERT Solutions Class 7
• NCERT Solutions Class 8
• NCERT Solutions Class 9
• NCERT solutions Class 10
• NCERT solutions Class 11
• NCERT solutions Class 12

#### NCERT Exemplar for Class 9 Mathematics

NCERT Exemplar helps in laying the foundation to all the basic as well as advanced concepts in such a way that the answers are self-explanatory, meaning students may not always have to depend on  teachers to clarify their doubts while studying, especially during the last minute preparation making  it easier to understand the concepts quickly and thoroughly.   NCERT Exemplar has a repository of  NCERT related questions. As a result, teachers and mentors recommend  students to include NCERT Exemplar books as an integral part of  their study material. It is  a great guide for students to step up their preparation to get excellent results.

The book is specially designed by subject matter experts, explained in an easy to understand language, all the theorems, formulas and exercises covered in e NCERT Class 9 Mathematics textbook. You can get the NCERT Exemplar for Class 9 Mathematics  from the Extramarks’ website.

Students can find questions ranging from basic to advanced level; thus, they become capable of solving all the levels of questions. After referring to Mathematics Class 9 Chapter 8 and NCERT Exemplar, students can be rest assured  that nothing remains untouched and every example solution, exercise  has been covered in the chapter and hence they are  confident of their preparation and ace the exam with excellent results.

#### Key Features of NCERT Solutions for Class 9 Mathematics Chapter 8

A strong mindset always gives rise to better results. Hence, we have prepared NCERT Solutions for Class 9 Mathematics Chapter 8 in such a way that helps students to develop analytical  mindset.   Some of the key features are:

• The academic notes are designed in a way that helps students to manage their time efficiently
• Students learn to quickly grasp concepts, formulas, definitions, theorems, proofs , postulates and calculations. The in-text and  end- text questions in the chapter , students will get enough practice to improvise their mathematical skills and enjoy solving mathematical problems.
• It enhances  the confidence of the students and hence they  will be able to leverage their performance.Understanding concepts  would be easier if you have followed NCERT Solutions to clarify your concepts to solve long and short answer questions, MCQs, and intext and end text questions. Also, don’t forget to take notes and practise with sample papers. will make it easier to grasp topics  mathematical skills
•  The students will be able to interpret proofs and draw conclusions to all  geometrical applications easily.

Q.1 The angles of quadrilateral are in the ratio 3 : 5 : 9 : 13. Find all the angles of the quadrilateral.

Ans.

Let four angles of quadrilateral be 3x, 5x, 9x and 13x. Then, by angle sum property in quadrilateral

3x + 5x + 9x + 13x = 360°

30x = 360°
x = 360°/30
= 12°

So, the first angle of quadrilateral
= 3(12°)
= 36°

= 5(12°)
= 60°

= 9(12°)
= 108°

= 13(12°)
= 156°

Q.2 If the diagonals of a parallelogram are equal, then show that it is a rectangle.

Ans. $\begin{array}{l}\text{Given: Let ABCD be a parallelogram, in which AC}=\text{BD.}\\ \text{To prove: ABCD is a rectangle.}\\ \text{Proof: In ΔADC and ΔBCD},\\ \text{\hspace{0.17em}AD}=\text{BC\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\text{Opposite sides of parallelogram are equal.}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}\mathrm{DC}=\mathrm{CD}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{Common}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{A}\mathrm{C}=\mathrm{BD}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\text{Given}\right]\\ \therefore \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{\Delta ADC}\cong \mathrm{\Delta BCD}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{By}\text{S.S.S.}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{ADC}=\angle \mathrm{BCD}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{By}\text{C.P.C.T.}\right]\\ \text{but\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{ADC}+\angle \mathrm{BCD}=180\mathrm{°}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{Co}-\text{interior angles}\right]\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{ADC}=\angle \mathrm{BCD}=90\mathrm{°}\\ \text{Since},\text{one angle of parallelogram is 90°.}\\ \text{So, ABCD is a rectangle.\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Hence proved.}\end{array}$

Q.3 Show that if the diagonals of a quadrilateral bisect each other at right angles, then it is a rhombus.

Ans. $\begin{array}{l}\text{Given}:\text{Let ABCD be a quadrilateral},\text{in which AO}=\mathrm{OC},\text{}\\ \text{}\mathrm{BO}=\mathrm{OD}.\text{}\mathrm{AC}\text{and BD bisect each other at 9}0\mathrm{°}.\\ \text{To prove}:\text{ABCD is a rhombus}.\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Proof}:\text{​\hspace{0.17em}\hspace{0.17em}In\hspace{0.17em}}\mathrm{\Delta AOB}\text{and}\mathrm{\Delta COB}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{AO}=\mathrm{OC}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}\left[\mathrm{Given}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{AOB}=\angle \mathrm{COB}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{Each}\text{90°}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{OB}=\mathrm{OB}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{Common}\right]\\ \therefore \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{\Delta AOB}\cong \mathrm{\Delta COB}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{By}\text{SAS}\right]\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{AB}=\mathrm{BC}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\left(\mathrm{i}\right)\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{By}\text{C.P.C.T.}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}In\hspace{0.17em}}\mathrm{\Delta BOC}\text{and}\mathrm{\Delta DOC}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{BO}=\mathrm{OD}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{Given}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{BOC}=\angle \mathrm{DOC}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{Each}\text{90°}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{OC}=\mathrm{OC}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}\left[\text{Common}\right]\\ \therefore \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{BOC}\cong \mathrm{\Delta DOC}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{By}\text{SAS}\right]\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{BC}=\mathrm{DC}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\left(\mathrm{ii}\right)\text{\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{By}\text{C.P.C.T.}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}In\hspace{0.17em}}\mathrm{\Delta COD}\text{and}\mathrm{\Delta DOA}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{CO}=\mathrm{OA}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{Given}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{COD}=\angle \mathrm{AOD}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{Each}\text{90°}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{OD}=\mathrm{OD}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\text{Common}\right]\\ \therefore \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{\Delta COD}\cong \mathrm{\Delta DOA}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{By}\text{SAS}\right]\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{DC}=\mathrm{DA}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\left(\mathrm{iii}\right)\left[\mathrm{By}\text{C.P.C.T.}\right]\\ \text{From equation}\left(\mathrm{i}\right),\left(\mathrm{ii}\right)\text{and}\left(\mathrm{iii}\right),\text{we get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}AB}=\mathrm{BC}=\mathrm{CD}=\mathrm{DA}\\ \text{Since},\text{all sides of quadrilateral ABCD are equal, so}\\ \text{ABCD is a rhombus.\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Hence proved.}\end{array}$

Q.4 Show that the diagonals of a square are equal and bisect each other at right angles.

Ans. $\begin{array}{l}\text{Given: Let ABCD be a square.}\\ \mathrm{To}\text{prove:AC}=\text{BD, A}\mathrm{O}=\mathrm{OC},\text{}\mathrm{BO}=\text{OD and AC}\perp \text{BD.}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Proof: In}\mathrm{\Delta ABC}\text{and}\mathrm{\Delta DCB}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{AB}=\mathrm{DC}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\text{Sides of square}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{ABC}=\angle \mathrm{DCB}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\text{Each 90°}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{BC}=\mathrm{BC}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\text{Common]}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{\Delta ABC}\cong \mathrm{\Delta DCB}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{By}\text{S.A.S.}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{AC}=\mathrm{BD}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{By}\text{C.P.C.T.}\right]\\ \text{In}\mathrm{\Delta AOB}\text{and}\mathrm{\Delta COD}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{AB}=\mathrm{CD}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\text{Sides of square}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{AOB}=\angle \mathrm{COD}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\text{Vertical opposite angles}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{BAO}=\angle \mathrm{DCO}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\text{Alternate interior angles}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{\Delta AOB}\cong \mathrm{\Delta COD}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{By}\text{S.A.S.}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{AO}=\mathrm{OC}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{By}\text{C.P.C.T.}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{BO}=\mathrm{OD}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{By}\text{C.P.C.T.}\right]\\ \text{In}\mathrm{\Delta AOB}\text{and}\mathrm{\Delta COB}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{AB}=\mathrm{CB}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\text{Sides of square}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{OB}=\mathrm{OB}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\text{Common}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{AO}=\mathrm{OC}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\text{Proved above}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{\Delta AOB}\cong \mathrm{\Delta COB}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{By}\text{S.A.S.}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{AOB}=\angle \mathrm{COB}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{By}\text{C.P.C.T.}\right]\\ \mathrm{and}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{AOB}+\angle \mathrm{COB}=180\mathrm{°}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{AOB}=90\mathrm{°}\\ \text{Thus},\text{diagonals AC and BD are equal and bisect each}\\ \text{other at 90°.\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Hence proved.}\end{array}$

Q.5 Show that if the diagonals of a quadrilateral are equal and bisect each other at right angles, then it is a square.

Ans. $\begin{array}{l}\text{Given}:\mathrm{Let}\text{ABCD be a quadrilateral. AC}=\text{BD and AC}\perp \text{BD,}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}OA}=\text{OC and OB}=\text{OD.}\\ \text{To prove: ABCD is a square.}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Proof:In}\mathrm{\Delta AOB}\text{and}\mathrm{\Delta COB}\\ \text{\hspace{0.17em}\hspace{0.17em}}\mathrm{A}\mathrm{O}=\mathrm{OC}\left[\mathrm{Given}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{AOB}=\angle \mathrm{COB}\text{}\left[\mathrm{Each}\text{90°}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{OB}=\mathrm{OB}\left[\mathrm{Common}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{\Delta AOB}\cong \mathrm{\Delta COB}\left[\mathrm{By}\text{S.A.S.}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{AB}=\mathrm{BC}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\left(\mathrm{i}\right)\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{By}\text{C.P.C.T.}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}In}\mathrm{\Delta BOC}\text{and}\mathrm{\Delta COD}\\ \text{\hspace{0.17em} \hspace{0.17em}}\mathrm{BO}=\mathrm{OD}\left[\mathrm{Given}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{BOC}=\angle \mathrm{DOC}\left[\mathrm{Each}\text{90°}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{OC}=\mathrm{OC}\left[\mathrm{Common}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{\Delta BOC}\cong \mathrm{\Delta COD}\left[\mathrm{By}\text{S.A.S.}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{BC}=\mathrm{CD}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\left(\mathrm{ii}\right)\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{By}\text{C.P.C.T.}\right]\\ \text{In}\mathrm{\Delta COD}\text{and}\mathrm{\Delta DOA}\\ \text{\hspace{0.17em} \hspace{0.17em}}\mathrm{CO}=\mathrm{OA}\left[\mathrm{Given}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{DOC}=\angle \mathrm{DOA}\left[\mathrm{Each}\text{90°}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{OD}=\mathrm{OD}\left[\mathrm{Common}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{\Delta COD}\cong \mathrm{\Delta DOA}\left[\mathrm{By}\text{S.A.S.}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{CD}=\mathrm{DA}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\left(\mathrm{iii}\right)\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{By}\text{C.P.C.T.}\right]\\ \text{From equation}\left(\mathrm{i}\right),\left(\mathrm{ii}\right)\text{and}\left(\mathrm{iii}\right),\text{we have}\\ \mathrm{AB}=\mathrm{BC}=\mathrm{CD}=\mathrm{DA}\\ \text{Thus},\text{ABCD is a square.Hence proved.}\end{array}$

Q.6 Diagonal AC of a parallelogram ABCD bisects ∠A . Show that
(i) it bisects ∠C and
(ii) ABCD is a rhombus. Ans.

$\begin{array}{l}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Given}:\text{ABCD is a parallelogram in which diagonal AC}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}bisects}\angle \text{A.}\\ \text{To prove}:\left(\mathrm{i}\right)\text{}\mathrm{it}\text{bisects\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{C}\text{also},\\ \text{\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}\left(\mathrm{ii}\right)\text{}\mathrm{ABCD}\text{}\mathrm{is}\text{}\mathrm{a}\text{rhombus}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Proof:}\left(\mathrm{i}\right)\text{Since, AD}\parallel \text{BC}\\ \text{so,}\angle \mathrm{DAC}=\angle \mathrm{BCA}\text{}...\left(\mathrm{i}\right)\left[\text{Alternate interior angles}\right]\\ \mathrm{and}\text{AB}\parallel \text{DC}\\ \text{so,\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{BAC}=\angle \mathrm{DCA}...\left(\mathrm{ii}\right)\left[\text{Alternate interior angles}\right]\\ \mathrm{But}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{BAC}=\angle \mathrm{DAC}\text{\hspace{0.17em}\hspace{0.17em}}...\left(\mathrm{iii}\right)\left[\mathrm{Given}\right]\\ \mathrm{So},\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{DCA}=\angle \mathrm{BCA}\left[\text{From equation}\left(\mathrm{i}\right)\text{and}\left(\mathrm{ii}\right)\right]\\ ⇒\mathrm{AC}\text{bisects}\angle \mathrm{C}.\\ \text{(ii) From equation}\left(\mathrm{i}\right)\text{and equation}\left(\mathrm{iii}\right),\text{we have}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{BAC}=\angle \mathrm{BCA}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{BC}=\mathrm{BA}\left[\begin{array}{l}\text{Sides opposite to equal angles}\\ \text{are equal.}\end{array}\right]\\ \text{Since},\text{adjacent sides of parallelogram are equal, so ABCD}\\ \text{is a rhombus. Hence proved.}\end{array}$

Q.7 ABCD is a rhombus. Show that diagonal AC bisects ∠A as well as ∠C and diagonal BD bisects ∠B as well as ∠D.

Ans. $\begin{array}{l}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Given}:\mathrm{ABCD}\text{is a rhombus.}\\ \text{To prove: AC bisects}\angle \text{A as well as}\angle \text{C.}\\ \text{\hspace{0.17em}\hspace{0.17em}BD bisects}\angle \text{B as well as}\angle \text{D.}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Proof:In}\mathrm{\Delta ABC},\text{AB}=\text{BC}\\ \text{So,}\angle \text{BAC}=\angle \text{BCA}...\left(\mathrm{i}\right)\left[\begin{array}{l}\mathrm{Opposite}\text{angles of equal sides}\\ \text{are equal.}\end{array}\right]\\ \text{Since,}\mathrm{AB}\parallel \mathrm{DC}\\ \text{So},\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \text{BAC}=\angle \text{DCA}...\left(\mathrm{ii}\right)\left[\text{Alternate interior angles.}\right]\\ \text{From equation}\left(\mathrm{i}\right)\text{and equation}\left(\mathrm{ii}\right),\text{we have}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \text{BCA}=\angle \text{DCA}\\ ⇒\mathrm{diagonal}\text{AC bisects}\angle \text{C.}\\ \text{Since, AD∥BC}\\ \mathrm{So},\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \text{DAC}=\angle \text{BCA}\dots \left(\mathrm{iii}\right)\left[\text{Alternate interior angles.}\right]\\ \text{From equation}\left(\mathrm{i}\right)\text{and equation}\left(\mathrm{iii}\right),\text{we have}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \text{BAC}=\angle \text{DAC}\\ ⇒\text{diagonal AC bisects}\angle \text{A.}\\ \text{Thus, AC bisects}\angle \text{A as well as}\angle \text{C.}\\ \text{Similarly, BC bisects}\angle \text{B as well as}\angle \text{D.Hence proved.}\end{array}$

Q.8 ABCD is a rectangle in which diagonal AC bisects ∠A as well as ∠C. Show that:
(i) ABCD is a square.
(ii) diagonal BD bisects ∠B as well as ∠D.

Ans. $\begin{array}{l}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Given}:\mathrm{ABCD}\text{is a rectangle. Diagonal}\mathrm{AC}\text{bisects}\angle \mathrm{A}\text{as well}\\ \text{\hspace{0.17em}\hspace{0.17em}as}\angle \mathrm{C}.\\ \text{To prove:}\left(\mathrm{i}\right)\text{ABCD is a square.}\\ \left(\mathrm{ii}\right)\text{Diagonal}\mathrm{BD}\text{bisects}\angle \mathrm{B}\text{as well as}\angle \mathrm{D}.\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Proof:}\left(\mathrm{i}\right)\text{Since,}\angle 1=\angle 2\text{and}\angle 3=\angle 4\\ \because \mathrm{AB}\parallel \mathrm{CD}\\ \therefore \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle 2=\angle 3\text{}\left[\text{Alternate interior angles}\right]\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle 1=\angle 3\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{CD}=\mathrm{AD}\left[\begin{array}{l}\text{Opposite sides of equal angles}\\ \text{are equal.}\end{array}\right]\\ \end{array}$

$\begin{array}{l}\mathrm{Since},\mathrm{adjecent}\text{sides of a rectangle are equal, so ABCD is}\\ \text{a square.}\\ \left(\mathrm{ii}\right)\mathrm{As}\text{ABCD is a square.}\\ \text{So, \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{AB}=\mathrm{AD}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{ADB}=\angle \mathrm{ABD}...\left(\mathrm{i}\right)\left[\begin{array}{l}\mathrm{Opposite}\text{angles of equal sides}\\ \text{are equal.}\end{array}\right]\\ \mathrm{Since},\mathrm{AB}\parallel \mathrm{CD}\\ \mathrm{So},\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{ABD}=\angle \mathrm{CDB}...\left(\mathrm{ii}\right)\left[\mathrm{Alternate}\text{interior angles}\right]\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{ADB}=\angle \mathrm{CDB}\end{array}$

$\begin{array}{l}\end{array}$ $\begin{array}{l}⇒\mathrm{BD}\text{bisects}\angle \mathrm{D}.\\ \text{Similarly},\text{we can prove that BD bisects}\angle \mathrm{B}.\\ \text{Therefore},\text{BD bisects}\angle \mathrm{B}\text{as well as}\angle \mathrm{D}.\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Hence proved.}\end{array}$

Q.9

$\begin{array}{l}\text{In parallelogram ABCD},\text{two points P and Q are taken on diagonal BD such that DP}=\text{BQ}.\text{Show that}:\\ \left(\mathrm{i}\right)\text{ }\mathrm{\Delta APD}\cong \mathrm{\Delta CQB}\\ \left(\mathrm{ii}\right)\text{ }\mathrm{AP}=\mathrm{CQ}\mathrm{}\\ \left(\mathrm{iii}\right)\text{ }\mathrm{\Delta AQB}\cong \mathrm{\Delta CPD}\mathrm{}\\ \left(\mathrm{iv}\right)\text{ }\mathrm{AQ}=\mathrm{CP}\mathrm{}\\ \left(\mathrm{v}\right)\text{ }\mathrm{APCQ}\text{is a parallelogram}\end{array}$ Ans.

$\begin{array}{l}\text{Given}:\mathrm{ABCD}\text{is a parallelogram and DP}=\text{BQ.}\\ \text{To prove:}\mathrm{}\left(\mathrm{i}\right)\mathrm{\Delta APD}\cong \mathrm{\Delta CQB}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}\left(\mathrm{ii}\right)\text{ }\mathrm{AP}=\mathrm{CQ}\mathrm{}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}\left(\mathrm{iii}\right)\text{ }\mathrm{\Delta AQB}\cong \mathrm{\Delta CPD}\mathrm{}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left(\mathrm{iv}\right)\text{ }\mathrm{AQ}=\mathrm{CP}\mathrm{}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left(\mathrm{v}\right)\text{ }\mathrm{APCQ}\text{is a parallelogram}\\ \text{\hspace{0.17em}Proof:\hspace{0.17em}In}\mathrm{\Delta APD}\text{}\mathrm{and}\text{}\mathrm{\Delta CQB}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{AD}=\mathrm{BC}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\text{Opposite sides of parallelogram.}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{ADP}=\angle \mathrm{CBQ}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}\left[\text{Alternate}\mathrm{r}\text{interior angles.}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{DP}=\mathrm{BQ}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\text{Given}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\therefore \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{\Delta APD}\cong \mathrm{\Delta CQB}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{By}\text{S.A.S.}\right]\\ \left(\mathrm{ii}\right)\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{AP}=\mathrm{CQ}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}\left[\mathrm{By}\text{C.P.C.T.}\right]\\ \left(\mathrm{iii}\right)\text{In}\mathrm{\Delta AQB}\text{}\mathrm{and}\text{}\mathrm{\Delta CPD}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{AB}=\mathrm{CD}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{Opposite}\text{sides of parallelogram.}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{ABQ}=\angle \mathrm{CDP}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{Alternater}\text{interior angles.}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{BQ}=\mathrm{DP}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}\left[\mathrm{Given}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\therefore \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{\Delta AQB}\cong \mathrm{\Delta CPD}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{By}\text{S.A.S.}\right]\\ \left(\mathrm{iv}\right)\text{ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{AQ}=\mathrm{PC}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}\left[\mathrm{By}\text{C.P.C.T.}\right]\\ \left(\mathrm{v}\right)\text{ }\mathrm{Since},\text{AP}=\text{CQ and AQ}=\text{PC}\\ \text{So, APCQ is a parallelogram.}\text{[Opposite sides are parallel.}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{Hence}\text{proved.}\end{array}$

Q.10

$\begin{array}{l}\text{ABCD is a parallelogram and AP and CQ are perpendiculars}\\ \text{from vertices A and C on diagonal BD }\left(\text{see the figure below}\right)\text{.}\\ \text{Show that}\\ \text{\hspace{0.17em}}\left(\text{i}\right)\text{ ΔAPB }\cong \text{ ΔCQD}\\ \left(\text{ii}\right)\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}AP=CQ}\end{array}$ Ans.

$\begin{array}{l}\text{Given:ABCD is a parallelogram and AP }\perp \text{ BD and CQ }\perp \text{ BD.}\\ \text{To prove : }\left(\text{i}\right)\text{ ΔAPB }\cong \text{ ΔCQD}\\ \left(\text{ii}\right)\text{ AP = CQ}\\ \text{Proof : }\left(\text{i}\right)\text{\hspace{0.17em}In ΔAPB\hspace{0.17em}and\hspace{0.17em}ΔCQD}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \text{ABP = }\angle \text{CDQ\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\text{Alternate interior angles}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \text{APB = }\angle \text{CQD\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}[Each 90°]}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}AB = CD\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\text{Opposite sides of parallelogram.}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\therefore \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}ΔAPB }\cong \text{ ΔCQD\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\text{By A.A.S.}\right]\\ \left(\text{ii}\right)\text{ AP = CQ\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\text{By C.P.C.T.}\right]\text{ Hence proved.}\end{array}$

Q.11

$\begin{array}{l}\text{In}\mathrm{\Delta }\text{ABC and}\mathrm{\Delta }\text{DEF},\text{AB}=\text{DE},\text{AB}||\text{DE},\text{BC}=\text{EF and BC}||\text{EF}.\text{}\\ \text{Vertices A},\text{B and C are joined to vertices D},\text{E and F respectively}\left(\text{see the figure below}\right).\text{Show that}\\ \left(\mathrm{i}\right)\mathrm{quadrilateral}\mathrm{ABED}\mathrm{is}\mathrm{a}\mathrm{parallelogram}\\ \left(\mathrm{ii}\right)\mathrm{quadrilateral}\mathrm{BEFC}\mathrm{is}\mathrm{a}\mathrm{parallelogram}\\ \left(\mathrm{iii}\right)\mathrm{AD}||\mathrm{CF}\mathrm{and}\mathrm{AD}=\mathrm{CF}\\ \left(\mathrm{iv}\right)\mathrm{quadrilateral}\mathrm{ACFD}\mathrm{is}\mathrm{a}\mathrm{parallelogram}\\ \left(\mathrm{v}\right)\text{ }\mathrm{AC}=\mathrm{DF}\\ \left(\mathrm{vi}\right)\text{ }\mathrm{\Delta ABC}\cong \mathrm{\Delta DEF}.\end{array}$ Ans.

$\begin{array}{l}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Given}:\mathrm{In}\text{}\mathrm{\Delta ABC}\text{and}\mathrm{\Delta DEF},\text{AB}=\text{DE},\text{AB}||\text{DE},\mathrm{}\text{BC}=\text{EF and}\\ \text{BC}||\text{EF.}\\ \text{To prove:}\left(\mathrm{i}\text{)quadrilateral}\mathrm{ABED}\mathrm{is}\mathrm{a}\text{parallelogram}\\ \left(\mathrm{ii}\right)\text{ quadrilateral}\mathrm{BEFC}\mathrm{is}\mathrm{a}\text{parallelogram}\\ \left(\mathrm{iii}\right)\text{ }\mathrm{AD}||\mathrm{CF}\mathrm{and}\mathrm{AD}=\mathrm{CF}\\ \left(\mathrm{iv}\right)\text{ quadrilateral A}\mathrm{CFD}\mathrm{is}\mathrm{a}\text{parallelogram}\\ \left(\mathrm{v}\right)\text{ }\mathrm{AC}=\mathrm{DF}\\ \left(\mathrm{vi}\right)\text{ }\mathrm{\Delta ABC}\cong \mathrm{\Delta DEF}.\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Proof:}\left(\mathrm{i}\right)\text{Since, AB}=\text{DE and AB}\parallel \text{DE}\\ \text{So, ABED is a parallelogram.}\\ \left[\mathrm{One}\text{pair of opposite sides is parallel and equal.}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}(ii) Since},\text{BC}=\text{EF and BC}||\text{EF}\\ \text{So, BEFC is a parallelogram.}\\ \left[\text{One pair of opposite sides is parallel and equal.}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}(iii) Since, ABED is a prallelogram.}\\ \text{So, AD}=\text{BE and AD}\parallel \text{BE\hspace{0.17em}\hspace{0.17em}}...\left(\mathrm{i}\right)\\ \left[\mathrm{One}\text{pair of opposite sides is parallel and equal.}\right]\\ \text{Since},\text{BEFC is a prallelogram.}\\ \text{So, BE}=\text{CF and BE}\parallel \text{CF\hspace{0.17em}\hspace{0.17em}}...\left(\mathrm{ii}\right)\\ \left[\text{One pair of opposite sides is parallel and equal.}\right]\\ \text{From equation}\left(\mathrm{i}\right)\text{and equation}\left(\mathrm{ii}\right),\text{we have}\\ \text{AD}=\text{CF and AD}\parallel \text{CF}\\ \text{(iv) Since,\hspace{0.17em}\hspace{0.17em}AD}=\text{CF and AD}\parallel \text{CF}\\ \text{So},\text{ADFC is a prallelogram.}\\ \left[\text{One pair of opposite sides is parallel and equal.}\right]\\ \text{(v) Since},\text{ADFC is a prallelogram.}\\ \text{So, AC}=\text{DF}\left[\text{Opposite sides of parallelogram.}\right]\\ \left(\mathrm{vi}\right)\text{ In}\mathrm{\Delta ABC}\text{and}\mathrm{\Delta DEF}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{AB}=\mathrm{DE}\left[\text{Given}\right]\\ \mathrm{BC}=\mathrm{EF}\left[\text{Given]}\\ \mathrm{AC}=\mathrm{DF}\left[\text{Proved above}\right]\\ \therefore \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{\Delta ABC}\cong \mathrm{\Delta DEF}\left[\mathrm{By}\text{\hspace{0.17em}}\mathrm{S}.\mathrm{S}.\mathrm{S}.\right]\mathrm{Hence}\text{proved.}\end{array}$

Q.12

$\begin{array}{l}\mathrm{ABCD}\mathrm{is}\mathrm{a}\mathrm{trapezium}\mathrm{in}\mathrm{which}\mathrm{AB}||\mathrm{CD}\mathrm{and}\mathrm{AD}=\mathrm{BC}\\ \left(\mathrm{see}\mathrm{the}\mathrm{figure}\mathrm{below}\right).\mathrm{Show}\mathrm{that}\\ \left(\mathrm{i}\right)\text{ }\angle \mathrm{A}=\angle \mathrm{B}\\ \left(\mathrm{ii}\right)\text{ }\angle \mathrm{C}=\angle \mathrm{D}\\ \left(\mathrm{iii}\right)\text{ }\mathrm{\Delta ABC}\cong \mathrm{\Delta BAD}\\ \left(\mathrm{iv}\right)\text{ }\mathrm{diagonal}\mathrm{AC}=\mathrm{diagonal}\mathrm{BD}\end{array}$ Ans. $\begin{array}{l}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{Given}:\mathrm{In}\text{trapezium ABCD, AB}\parallel \text{CD and AD}=\text{BC.}\\ \text{To prove:}\left(\mathrm{i}\right)\angle \mathrm{A}=\angle \mathrm{B}\\ \text{\hspace{0.17em}\hspace{0.17em}}\left(\mathrm{ii}\right)\angle \mathrm{C}=\angle \mathrm{D}\\ \left(\mathrm{iii}\right)\mathrm{\Delta ABC}\cong \mathrm{\Delta BAD}\\ \left(\mathrm{iv}\right)\text{diagona}\mathrm{l}\mathrm{AC}=\text{diagonal}\mathrm{BD}\\ \text{Construction}:\text{Draw CE}\parallel \text{DA.\hspace{0.17em}Join AC.}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Proof:Since, AB}\parallel \mathrm{CD}\text{}⇒\text{AE}\parallel \mathrm{CD}\\ \text{and CE}\parallel \mathrm{DA}\\ \text{So},\text{AECD is a prallelogram.}\\ \text{Then, AD}=\text{CE}\\ \text{But, AD}=\text{BC}\\ ⇒\mathrm{CE}=\mathrm{BC}\end{array}$

$\begin{array}{l}⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{CBE}=\angle \mathrm{CEB}\left[\begin{array}{l}\mathrm{Opposite}\text{sides of equal angles are}\\ \text{equal in}\mathrm{\Delta }\text{CEB.}\end{array}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{CBE}+\angle \mathrm{CBA}=180\mathrm{°}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\left(\mathrm{i}\right)\left[\mathrm{Linear}\text{pair of angles}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{CEA}+\angle \mathrm{DAE}=180\mathrm{°}\text{\hspace{0.17em}\hspace{0.17em}}\left[\text{Cointerior angles}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{CBE}+\angle \mathrm{DAE}=180\mathrm{°}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\left(\mathrm{ii}\right)\left[\because \angle \mathrm{CBE}=\angle \mathrm{CEA}\right]\\ \mathrm{From}\text{equation}\left(\mathrm{i}\right)\text{and equation\hspace{0.17em}\hspace{0.17em}}\left(\mathrm{ii}\right),\text{we have}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{CBE}+\angle \mathrm{CBA}=\angle \mathrm{CBE}+\angle \mathrm{DAE}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{CBA}=\angle \mathrm{DAE}\\ ⇒\angle \mathrm{B}=\angle \mathrm{A}\\ ⇒\angle \mathrm{A}=\angle \mathrm{B}\\ \left(\mathrm{ii}\right)\mathrm{S}\mathrm{ince},\text{AB}\parallel \text{DC}\\ \text{So,\hspace{0.17em}}\angle \mathrm{A}+\angle \mathrm{D}=180\mathrm{°}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{and}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{B}+\angle \mathrm{C}=180\mathrm{°}\\ ⇒\angle \mathrm{A}+\angle \mathrm{D}=\angle \mathrm{B}+\angle \mathrm{C}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{D}=\angle \mathrm{C}\left[\because \angle \mathrm{A}=\angle \mathrm{B}\right]\\ \text{(iii) In}\mathrm{\Delta ABC}\text{and}\mathrm{\Delta BAD}\\ \mathrm{AB}=\mathrm{BA}\left[\mathrm{Common}\right]\\ \angle \mathrm{A}=\angle \mathrm{B}\left[\mathrm{Proved}\text{above}\right]\\ \mathrm{AD}=\mathrm{BC}\left[\mathrm{Given}\right]\\ \therefore \mathrm{\Delta ABC}\cong \mathrm{\Delta BAD}\left[\mathrm{By}\text{S.A.S.}\right]\\ \left(\mathrm{iv}\right)\text{}\mathrm{AC}=\mathrm{BD}\left[\mathrm{By}\text{C.P.C.T.}\right]\\ \text{Thus},\text{diagonal AC = diagonal BD.Hence proved.}\end{array}$

Q.13 ABCD is a quadrilateral in which P, Q, R and S are mid-points of the sides AB, BC, CD and DA (see the figure below). AC is a diagonal. Show that :
(i) SR||AC and SR = (1/2)AC
(ii) PQ = SR
(iii) PQRS is a parallelogram. Ans.

$\begin{array}{l}\text{Given}:\text{ABCD is a quadrilateral in which P, Q, R and S \hspace{0.17em}are}\\ \text{\hspace{0.17em}mid-points of the sides AB, BC, CD and DA.}\\ \text{To prove:}\left(\text{i}\right)\text{SR||AC and SR}=\frac{\text{1}}{2}\text{AC}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left(\text{ii}\right)\text{PQ}=\text{SR}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left(\text{iii}\right)\text{PQRS is a parallelogram.}\\ \text{Proof: \hspace{0.17em}}\left(\mathrm{i}\right)\mathrm{I}\mathrm{n}\text{}\mathrm{\Delta ADC},\text{\hspace{0.17em}}\mathrm{S}\text{and R are the mid-points of DA and DC}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}respectively. So, by mid-point theorem}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}SR}\parallel \mathrm{AC}\text{and SR=}\frac{1}{2}\mathrm{AC}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\left(\mathrm{i}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}(ii) In}\mathrm{\Delta ABC},\text{\hspace{0.17em}}\mathrm{P}\text{and Q are the mid-points of AB and BC}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}respectively. So, by mid-point theorem}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}PQ}\parallel \mathrm{AC}\text{and PQ}=\frac{1}{2}\mathrm{AC}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\left(\mathrm{ii}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}From equation}\left(\mathrm{i}\right)\text{and equation}\left(\mathrm{ii}\right),\text{we get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}PQ}=\text{SR}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left(\mathrm{iii}\right)\mathrm{From}\text{equation}\left(\mathrm{i}\right)\text{and equaton}\left(\mathrm{ii}\right),\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}\mathrm{PQ}=\mathrm{SR}\text{and PQ}\parallel \mathrm{SR}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{PQSR}\text{is a prallelogram.}\left[\begin{array}{l}\text{One pair of opposite}\\ \text{sides is equal and parallel.}\end{array}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Hence proved.}\end{array}$

Q.14 ABCD is a rhombus and P, Q, R and S are the mid-points of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rectangle. Ans.

$\begin{array}{l}\text{Given}:\text{ABCD is a rhombus and P},\text{Q},\text{R and S are the}\mathrm{}\text{mid-points}\\ \text{of the sides AB},\text{BC},\text{CD and DA}\mathrm{}\text{respectively.}\\ \text{To prove:PQRS is a rectangle.}\\ \text{Construction}:\text{Draw AC and BD.}\\ \text{Proof: In}\mathrm{\Delta ABC},\text{P and Q are the mid-points of AB and BC}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}respectively.So, by mid-point theorem}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} PQ}=\frac{1}{2}\mathrm{AC}\text{and PQ}\parallel \mathrm{AC}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\left(\mathrm{i}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}In}\mathrm{\Delta ADC},\text{S and R are the mid-points of AD and DC}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}respectively. So, by mid-point theorem}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} SR}=\frac{1}{2}\mathrm{AC}\text{and SR}\parallel \mathrm{AC}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\left(\mathrm{ii}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}From equation(i) and equation}\left(\mathrm{ii}\right),\text{we have}\\ \text{\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}PQ}=\text{SR and PQ}\parallel \mathrm{SR}\\ \text{So, PQRS is a parallelogram.}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\begin{array}{l}\text{If one pair of opposite sides of a quadrilateral is parallel}\\ \text{and equal, then it is a parallelogram.}\end{array}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}In}\mathrm{\Delta ABD},\text{P and S are the mid-points of AB and AD}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}respectively. So, by mid-point theorem}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}PS}=\frac{1}{2}\mathrm{BD}\text{and PS}\parallel \mathrm{BD}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\left(\mathrm{iii}\right)\\ \text{From equation}\left(\mathrm{i}\right)\text{and equation}\left(\mathrm{iii}\right),\text{we have}\\ \text{PN}\parallel \mathrm{MO}\text{and MP}\parallel \mathrm{ON}\\ \text{So, PMON is a parallelogram.}\\ \text{}\angle \mathrm{P}=\angle \mathrm{MON}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\text{Opposite angles of parallelogram are equal.}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=90\mathrm{°}\\ \text{Since,}\angle \mathrm{P}=90\mathrm{°}\text{in parallelogram PQRS, so PQRS is a rectangle.}\end{array}$

Q.15 ABCD is a rectangle and P, Q, R and S are mid-points of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rhombus.

Ans. $\begin{array}{l}\text{Given}:\text{ABCD is a rectangle and P},\text{Q},\text{R and S are the}\\ \text{mid-points of the sides AB},\text{BC},\text{CD and DA}\text{respectively}\text{.}\\ \text{To prove:PQRS is a rhombus}\text{.}\\ \text{Construction}:\text{Draw AC and BD}\text{.They are the diagonals of the quadrilateral.}\\ \text{Proof: In}\Delta ABC,\text{P and Q are the mid-points of AB and BC}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{respectively}\text{. So, by mid-point theorem}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{PQ}=\frac{1}{2}AC\text{and PQ}\parallel AC\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\dots \left(i\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{In ΔADC},\text{S and R are the mid-points of AD and DC}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{respectively}\text{. So, by mid-point theorem}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{SR}=\frac{1}{2}AC\text{and SR}\parallel AC\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\dots \left(ii\right)\\ \text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{From equation(i) and equation}\left(ii\right),\text{we have}\\ \text{}\text{\hspace{0.17em}}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{PQ}=\text{SR and PQ}\parallel SR\\ \text{So, PQRS is a parallelogram}\text{.}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left[\begin{array}{l}\text{If one pair of opposite sides of a quadrilateral is parallel}\\ \text{and equal, then it is a parallelogram}\text{.}\end{array}\right]\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{In ΔABD},\text{P and S are the mid-points of AB and AD}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{respectively}\text{. So, by mid-point theorem}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{PS}=\frac{1}{2}BD\text{and PS}\parallel BD\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\dots \left(iii\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}Since},\text{AC}=\text{BD}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left[Diagonals\text{of rectangle are equal}\text{.}\right]\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}⇒\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}PQ=PS\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left[\frac{1}{2}AC=\frac{1}{2}BD\right]\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}Since},\text{adjecent sides of a parallelogram ABCD are equal}\text{.}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{So, PQRS is rhombus}\text{.}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}Hence proved}\text{.}\end{array}$

Q.16 ABCD is a trapezium in which AB || DC, BD is a diagonal and E is the mid-point of AD.
A line is drawn through E parallel to AB intersecting BC at F (see Fig. 8.30). Show that F is the mid-point of BC. Ans. $\begin{array}{l}\text{Given}:\text{}\mathrm{ABCD}\text{is a trapezium in which}AB||DC,\text{BD is a}\text{\hspace{0.17em}}\text{diagonal}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{and E is the mid}-\text{point of AD}\text{. A line through E is prallel}\\ \text{}\text{\hspace{0.17em}}\text{to AB}\text{.}\\ \text{To prove: F is the mid-point of BC i}\text{.e}\text{., BF}=\text{CF}\text{.}\\ \text{Proof: In ΔABD},\text{E is mid-point of AD and EO}\parallel AB\\ \text{So, by converse of mid-point theorem,}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{OD}=\text{OB}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{In ΔBCD},\text{O is mid-point of BD and EF}\parallel CD\\ \text{So, by converse of mid-point theorem,}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{CF}=\text{FB}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{Therefore, F is the mid-point of BC}\text{.}\text{\hspace{0.17em}}\text{\hspace{0.17em}Hence proved}\text{.}\end{array}$

Q.17 In a parallelogram ABCD, E and F are the mid-points of sides AB and CD respectively (see Fig. 8.31).
Show that the line segments AF and EC trisect the diagonal BD. Ans.

$\begin{array}{l}\text{Given}:\text{In parallelogram ABCD, E and F are the mid-points of}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}sides AB and CD respectively.}\\ \mathrm{To}\text{prove: DP}=\text{PQ}=\text{QB}\\ \text{Proof: In}\mathrm{\Delta DQC},\text{F is mid-point of DC and PF}\parallel \mathrm{QC},\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}So},\text{by converse of mid-point theorem, we have}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}DP}=\text{PQ\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\left(\mathrm{i}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}In ΔABP},\text{E is mid-point of AB and EQ}\parallel \mathrm{AP},\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{So},\text{by converse of mid-point theorem, we have}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}PQ}=\text{QB\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\left(\mathrm{ii}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{From}\text{equation}\left(\mathrm{i}\right)\text{and equation}\left(\mathrm{ii}\right),\text{we have}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}DP}=\text{PQ}=\mathrm{QB}\\ \text{This implies that line segments AF and EC trisect the diagonal BD}.\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Hence proved.}\end{array}$

Q.18 Show that the line segments joining the mid-points of the opposite sides of a quadrilateral bisect each other.

Ans. $\begin{array}{l}\text{Given}:\text{Let ABCD is a quadrilateral and P},\text{Q},\text{R and S are the}\\ \text{mid-points of the sides AB},\text{BC},\text{CD and DA}\text{respectively}\text{.}\\ \text{To prove:PR and QS bisect each other}\text{.}\\ \text{Construction}:\text{Draw AC}\text{.}\\ \text{Proof: In}\Delta ABC,\text{P and Q are the mid-points of AB and BC}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{respectively}\text{. So, by mid-point theorem}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{PQ}=\frac{1}{2}AC\text{and PQ}\parallel AC\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left(i\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{In}\Delta ADC,\text{S and R are the mid-points of AD and DC}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{respectively}\text{. So, by mid-point theorem}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{SR}=\frac{1}{2}AC\text{and SR}\parallel AC\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\dots \left(ii\right)\\ \text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{From equation(i) and equation}\left(ii\right),\text{we have}\\ \text{}\text{\hspace{0.17em}}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{PQ}=\text{SR and PQ}\parallel SR\\ \text{So, PQRS is a parallelogram}\text{.} \text{\hspace{0.17em}}\text{[}\begin{array}{l}\text{If one pair of opposite sides of a quadrilateral is parallel}\\ \text{and equal, then it is a parallelogram.}\end{array}\right]\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\\ \text{Since, diagonals of a parallelogram bisect each other, so}\\ \text{PR and QS bisect each other}\text{.}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}\hspace{0.17em}Hence proved}\text{.}\end{array}$

Q.19

$\begin{array}{l}\mathrm{ABC}\mathrm{is}\mathrm{a}\mathrm{triangle}\mathrm{right}\mathrm{angled}\mathrm{at}\mathrm{C}.\mathrm{A}\mathrm{line}\mathrm{through}\mathrm{the}\\ \mathrm{mid}–\mathrm{point}\mathrm{M}\mathrm{of}\mathrm{hypotenuse}\mathrm{AB}\mathrm{and}\mathrm{parallel}\mathrm{to}\mathrm{BC}\mathrm{intersects}\\ \mathrm{AC}\mathrm{at}D.\text{Show that}\\ \left(\mathrm{i}\right)\text{ }\mathrm{D}\text{is the mid-point of}\mathrm{AC}\\ \left(\mathrm{ii}\right)\text{ MD⊥AC}\\ \left(\mathrm{iii}\right)\text{ }\mathrm{CM}=\mathrm{MA}=\frac{1}{2}\mathrm{AB}\end{array}$

Ans. $\begin{array}{l}\text{Given:In\hspace{0.17em}}\mathrm{\Delta ABC},\text{}\angle \mathrm{C}=90\mathrm{°}\text{and M is mid-point of AB, MD}\parallel \mathrm{BC}.\\ \mathrm{To}\text{prove:}\left(\mathrm{i}\right)\text{D is the mid-point of AC}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}(ii)MD⊥AC}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}(iii)CM=MA=}\frac{1}{2}\mathrm{AB}\\ \text{Proof:}\left(\mathrm{i}\right)\text{In}\mathrm{\Delta ABC},\text{MD}\parallel \mathrm{BC}\text{and M is mid-point of AB, then}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} by converse of mid-point theorem,}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}AD}=\text{DC i.e.,}\mathrm{D}\text{is the mid-point of}\mathrm{AC}.\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left(\mathrm{ii}\right)\text{Since, MD}\parallel \mathrm{BC}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{So},\text{}\angle \mathrm{MDA}=\angle \mathrm{BCA}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\text{Corresponding angles}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=90\mathrm{°}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Therefore},\text{}\mathrm{MD}\perp \mathrm{AC}.\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left(\mathrm{iii}\right)\text{\hspace{0.17em}\hspace{0.17em}In\hspace{0.17em}ΔMDC and ΔMDA}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{DC}=\mathrm{DA}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}\left[\text{Proved above}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{MDC}=\angle \mathrm{MDA}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{Each}\text{90°\hspace{0.17em}as}\mathrm{MD}\perp \mathrm{AC}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}\mathrm{MD}=\mathrm{MD}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\text{Common]}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{\Delta MDC}\cong \mathrm{\Delta MDA}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{By}\text{S.A.S.}\right]\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}CM=MA\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{By}\text{C.P.C.T.}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Since},\text{AM}=\frac{1}{2}\mathrm{AB}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{So},\text{\hspace{0.17em}}\mathrm{CM}=\mathrm{MA}=\frac{1}{2}\mathrm{AB}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Hence proved.}\end{array}$