# NCERT Solutions for Class 10 Maths Chapter 6 Exercise 6.2

NCERT books are the main textbooks used by CBSE students, and with wise use, pupils can be well-prepared for their future endeavours. For students, NCERT books are quite beneficial and provide a range of advantages for increasing their foundational knowledge. The development of a deep comprehension of the fundamental ideas in every subject is supported by NCERT books. Students can thoroughly understand the concepts by attentively reading the NCERT textbooks. Long-term benefits include students being able to solve issues quickly and effectively, provided they acquire the essential conceptual clarity. If a student could understand a concept, there would be no need to memorise it. NCERT textbooks are exhaustive and all-inclusive in their own right, and CBSE rarely includes material from other sources in its exam questions. The pupils ought to read and study these works in depth. While there is nothing wrong with learning from non-NCERT books, students should make sure they have covered every single topic in those books. Before reading any other books, it is important to properly understand the NCERT texts.

Once they adhere to NCERT and rigorously study from these books, students will discover that they can rapidly respond to all of the questions from the previous years question papers. All of the questions in the CBSE Class 10 board test were sourced from NCERT. A small number of questions and phrases are manipulated to gauge the pupils’ knowledge. Students can get assistance from the NCERT Solutions For Class 10 Maths Chapter 6 Exercise 6.2 if they have any doubts. Students who intend to take competitive exams can use the NCERT Solutions For Class 10 Maths Chapter 6 Exercise 6.2. Students who plan to choose the science stream in Grade 11 should begin preparing in Class 10 itself, and NCERT Solutions For Class 10 Maths Chapter 6 Exercise 6.2 will be essential in this process. To dowell on the Class 10 exams, the students must become familiar with these NCERT solutions. The math specialists at Extramarks have answered these NCERT questions. These solutions will aid students in comprehending and mastering a range of Triangles-related issues. When trying to respond to questions in their annual exams or competitive exams, students who try to learn without comprehending may encounter difficulties. Students’ understanding of topics is a top emphasis in NCERT textbooks. In order to comprehend the questions and be able to respond to any topic, students must be able to apply the concepts from the NCERT texts. Understanding how to apply concepts to solve particular problems is crucial. Students must carefully observe the patterns given in the NCERT books in order to avoid becoming confused during the exam. A student who has a deep knowledge of questions can easily respond to any question that will be posed in exams. The NCERT Solutions For Class 10 Maths Chapter 6 Exercise 6.2 should be used to thoroughly study the material in order to ensure success.

The objective of the NCERT textbooks is for pupils to advance intellectually. NCERT Solutions have been put up by the Extramarks online learning platform to assist students in their academic endeavours while keeping in mind the format of NCERT books. Mathematics now used in Physics, Chemistry, Economics, engineering, and other fields. Mathematics necessary to completely understand various issues in these fields. Mathematics is practically everywhere, from the household to the art gallery. Mathematics includes things like forms, shapes, sizes, degrees, temperatures, money, time, and watches.Mathematics is taught as an “instrumental subject” in some curricula to help students learn other subjects, while integrated courses that combine Mathematics and other subjects are also available in other curricula. If students want to continue studying mathematics after class 10, they must have a stronger comprehension of the subject. For a deeper understanding, they can consult the NCERT Solutions For Class 10 Maths Chapter 6 Exercise 6.2.

## NCERT Solutions for Class 10 Maths Chapter 6 – Triangles

The NCERT Solutions for Class 10 Mathematics Chapter 6 Triangles, one of the most important study tools for CBSE Class 10 students, are provided here. The CBSE Syllabus for 2022-2023 is closely matched with Chapter 6 of the NCERT Solutions for Class 10 Mathematics.It contains several rules and theorems and covers a wide range of ground. When determining which theorem to apply to a particular question, students commonly become stuck. Each step required to resolve a problem is completely and clearly explained in the solutions provided by Extramarks.

Subject matter experts developed the NCERT Class 10 Maths Solutions to help students better prepare for their Class 10 board exams. These answers can be used to finish homework assignments and other tasks, as well as to get ready for tests. Students should constantly keep in mind that the NCERT Solutions For Class 10 Maths Chapter 6 Exercise 6.2 will come in handy anytime they run into a problem. To perform better in tests, they must adhere to the detailed solutions provided in the NCERT Solutions For Class 10 Maths Chapter 6 Exercise 6.2. Students who use NCERT Solutions For Class 10 Maths Chapter 6 Exercise 6.2 as a resource will therefore have no trouble answering the NCERT exercise questions.

### Importance of NCERT Solutions for Class 10 Maths Chapter 6 Exercise 6.2

The NCERT solutions to Class 10 Maths Ch 6 Ex 6.2 Triangles are available for students. The links provided below allow you to download the solutions in PDF format. These responses have been prepared with great clarity and precision, and they also include step-by-step explanations. A wide range of ideas pertaining to the triangle’s resemblance are covered in Exercise 6.2. If the corresponding angles and sides of two triangles are in the same proportion, then the triangles are said to be comparable. Additionally, two triangles are referred to as equiangular triangles if their corresponding angles are equal.

Students who have access to NCERT Solutions For Class 10 Maths Chapter 6 Exercise 6.2 will be able to answer all of the questions in this Exercise 6.2 that are based on related ideas. If pupils wish to comprehend an idea in its entirety, they must use NCERT Solutions. Students may find this chapter to be quite difficult, so they need to obtain the NCERT Solutions For Class 10 Maths Chapter 6 Exercise 6.2 to help them. Therefore, it is essential for students to have NCERT Solutions For Class 10 Maths Chapter 6 Exercise 6.2.

Students who get access to NCERT Solutions For Class 10 Maths Chapter 6 Exercise 6.2 will be able to solve all the questions based on similar concepts in this Exercise 6.2. NCERT Solutions are very important for students if they want to have a clear understanding of any concept. This chapter could be a little challenging for students, and that is why they must get the NCERT Solutions For Class 10 Maths Chapter 6 Exercise 6.2 to overcome those challenges. Hence, having NCERT Solutions For Class 10 Maths Chapter 6 Exercise 6.2 is crucial for students.

### Properties of Triangles

Before diving right into the Triangles chapter, students must first comprehend the properties of triangles. Triangles should be fundamentally understood by students because Class 10 may be too complicated for them. If they want to download the NCERT Solutions For Class 10 Maths Chapter 6 Exercise 6.2, they can always go to the Extramarks website. The first fundamental understanding a pupil must possess is that a triangle’s sides and angles serve as the foundation for all of its attributes. A triangle is a closed polygon with three sides and three vertices, according to the definition of a triangle.

The sum of the three interior angles of a triangle is 180 degrees. Based on the magnitude of the angles and the length of the sides, triangles can be classified into a wide variety of categories. The majority of subjects, including Classes 7, 8, 9, 10, and 11, cover the important concept of triangles. Students will learn about triangle properties, types, definitions, and mathematical significance in this section. NCERT Solutions For Class 10 Maths Chapter 6 Exercise 6.2 covers all of the chapter’s significant and minor details.

Students start to comprehend the properties of triangles, their types, and the theorems based on them, such as Pythagoras’ theorem, and others. Higher-level courses cover trigonometry, a subject where the right-angled triangle serves as the conceptual foundation. Students must download the PDF format of the NCERT Solutions for Class 10 Maths Chapter 6 Exercise 6.2 after understanding the triangle’s properties.

### Access NCERT Solutions for Class 10 Mathematics Chapter 6 – Triangles

The Extramarks website and mobile application offer all the resources needed for competitive examination preparation. Extramarks is a resource that students can use to prepare for a number of competitive exams, such as JEE Mains, NEET, JEE Advance, CUET, and others. Every chapter and subject’s solutions are available on Extramarks. For more help with that chapter, students can visit the Extramarks website and download the NCERT Solutions For Class 10 Maths Chapter 6 Exercise 6.2 in a PDF format. Students should be aware that every competitive test includes questions on fundamental Class 10 level mathematics, so in order to pass those exams easily, they should only focus on Class 10 mathematics using the NCERT Solutions for Class 10 Maths Chapter 6 Exercise 6.2.It is essential that students retain the entire syllabus of Class 10 to perform well on the Class 10 board examination. The scores students receive in Class 10 are significant because they will enable them to enrol in the chosen stream. The grades a student receives in Class 10 are crucial in determining their academic destiny. As a result, students must begin with a singular focus.

Students must be sincere while solving the questions, as understanding the concepts while solving the questions will help them in the long run. Students would be better prepared for challenging competitive examinations like the Joint Entrance Examination(JEE) with these basic Mathematics skills . Before jumping into any stream, students must have a clear idea in mind because that choice will affect their academic career. The NCERT Solutions For Class 10 Maths Chapter 6 Exercise 6.2 will aid students in developing a clear understanding. Hence, students must access the NCERT Solutions For Class 10 Maths Chapter 6 Exercise 6.2 from the website of Extramarks or the mobile application in PDF format for their own convenience.

### NCERT Chapter 6 Maths Class 10 Exercise 6.2

Questions based on the subject of Triangle Similarity are included in Exercise 6.2 of Class 10 NCERT Maths Chapter 6, Triangles. Exercise 6.2 for Class 10 consists of 10 Questions, 9 of which are Short Answer Questions and 1 is a Long Answer Question. The chapter involves a number of “Triangles”-related theorems and proofs, and Exercise 6.2 focuses mostly on 2 of them. The NCERT Solutions for Class 10 Maths, Chapter 6, Exercise 6.2, have covered all of the exercise’s questions.These theorems are used to demonstrate the triangles’ similarities. As a result, the following two theorems serve as the foundation for Chapter 6, the second exercise:

- Theorem 6.1: If a line is drawn parallel to one side of a triangle to intersect the other two sides at distinct points, the other two sides are divided in the same ratio.
- Theorem 6.2: If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.

Every theorem and answer of exercise 6.2 are available in the NCERT Solutions For Class 10 Maths Chapter 6 Exercise 6.2. Students can access the NCERT Solutions For Class 10 Maths Chapter 6 Exercise 6.2 to have a better understanding of the theorems and exercises.

The NCERT Solutions For Class 10 Maths Chapter 6 Exercise 6.2 is the perfect learning guide for students because NCERT Solutions are among the best study guides a student may have in order to get high marks on the annual examination. Questions from the NCERT textbook are frequently asked by CBSE, either directly or indirectly. As a result, students who understand concepts and problem-solving techniques in depth can easily achieve remarkable results on the Class 10 board examination.

### Triangles Class 10 Exercise 6.2

CBSE students of Class 10 can easily download the NCERT Solutions For Class 10 Maths Chapter 6 Exercise 6.2. They can download the NCERT Solutions For Class 10 Maths Chapter 6 Exercise 6.2 from the Extramarks website or mobile application in PDF format so that they can study the NCERT Solutions For Class 10 Maths Chapter 6 Exercise 6.2 even in offline mode.

Students encounter problems when completing the exercises in NCERT books and generally resolving queries. Students should practise answering the questions with confidence and make an effort to comprehend them. They can download the NCERT Solutions For Class 10 Maths Chapter 6 Exercise 6.2 for assistance. Students should remember that even though grades aren’t always taken into account, knowing Class 10 will assist them in passingthe competitive examinations. Out of all the subjects, Mathematics has the most significant impact on students’ lives. They must therefore concentrate on Mathematics. One of the most complex subjects is Mathematics, which is concept-based. Learning mathematics is good for the brain, and students should practise it frequently to improve their ability to think quickly and clearly. Mathematics has been an essential component of students’ development from the time they were young until they were in Class 10. Mathematics starts teaching students the fundamentals of addition, subtraction, multiplication, and division in Class 1 or Class 2, which will only help them when they start managing their finances. As a result, students must understand the concepts presented in NCERT Solutions for Class 10 Maths, Chapter 6, Exercise 6.2.

### Class 10 Maths Exercise 6.2 Solution

The NCERT Solutions For Class 10 Maths Chapter 6 Exercise 6.2 are the key material for students who are planning to appear in different competitions. NCERT books make the understanding of students quite clear so that they will be able to solve whatever questions may be asked in theirboard examinations or annual examinations. Therefore, Extramarks focuses on the NCERT books but provides solutions to a number of other books as well so that students will not face problems while solving questions. The CBSE rules and regulations are strictly followed by NCERT’s book.NCERT is the supreme body for management and research in the country. As a result, CBSE board students should be aware that CBSE examinations can be passed with high scores simply by studying the NCERT books.Extramarks serves as a learning platform for students to make solutions to all those questions that are in NCERT books and more. There is no doubt that the NCERT Solutions for Class 10 Maths, Chapter 6, Exercise 6.2, have been solved in order for students to score well and learn smart and simple steps in solving mathematics questions.Maths Class 10 Chapter 6 Exercise 6.2 provides ease for students to solve every question of the given exercise. Hence, it is true that students who study minutely and keenly will indeed score better in their board examinations, as they will have a crystal clear concept. Students may be hesitant to ask their teachers questions at times, but they can now find answers with the help of NCERT Solutions for Class 10 Maths Chapter 6 Exercise 6.2.Therefore, it is important for students to have a clear focus on and understanding of their academic careers.

### The Solution of Exercise 6.2 Class 10

Students must know that Extramarks is a platform where students are encouraged and helped by professional teachers of certain subjects. They must get access to Extramarks so that they can easily look over NCERT Solutions For Class 10 Maths Chapter 6 Exercise 6.2. That is why NCERT Solutions For Class 10 Maths Chapter 6 Exercise 6.2 are easily available for students so that they can rely on Extramarks for their doubts and queries.

### NCERT Solutions for Class 10 Maths

Extramarks offers resources for all boards, including CBSE, ICSE, and different state boards. Extramarks has made education easy for students. Extramarks has ensured that every chapter of Class 10 must be solved precisely and clearlyon this platform. Students can benefit greatly from the NCERT Solutions For Class 10 Maths Chapter 6 Exercise 6.2. Therefore, students are advised to look into Class 10 Maths Chapter 6 Exercise 6.2. Every question from the NCERT Class 10 Mathematics books has now been answered by Extramarks. Extramarks contains all of the answers, so students should begin new chapters with confidence.Extramarks’ primary goal is to teach students how to depend on themselves by simply illuminating the proper path. Every logical reasoning used to solve a problem is covered in NCERT Solutions for Class 10 Maths, Chapter 6, Exercise 6.2.

Question 1: The solution to question number 1 is based on the Basic Proportionality Theorem (B.P.T) or (Thales Theorem) Two triangles are similar if (i) Their corresponding angles are equal (ii) Their corresponding sides are in the same ratio (or proportion).

Question 2: The reasoning to solve question number 2 (i) is a line that divides any two sides of a triangle in the same ratio, then the line is parallel to the third side (converse of BPT). 2 (ii) a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side (converse of BPT). 2 (iii) a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side (converse of BPT).

Question 3: The reasoning to solve question 3 is if a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio.

Question 4: The reasoning to solve question 4 is if a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio.

Questions 5 and 6: The reasoning to solve question 5 is if a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side, and if a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.

Question 7: The reasoning to solve question 7 is that theorem 6.1 states that “If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio (BPT)”.

Question 8: The reasoning to solve question 8 is if a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side. (Converse of BPT).

Question 9: The reasoning to solve question 9 is if a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.

Question 10: The reasoning to solve question 10 is if a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.

Each question has been solved by NCERT Solutions For Class 10 Maths Chapter 6 Exercise 6.2. Therefore, students are advised to download the NCERT Solutions For Class 10 Maths Chapter 6 Exercise 6.2 to have a clear understanding of questions, concepts, and theorems.

Students can also download NCERT Solutions for other classes by clicking the links given below-

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### NCERT Solutions for Class 10 Maths Chapter 6 Exercises

Students can access the other exercises in NCERT Solutions for Class 10 Maths Chapter 6 Exercises as well.‘‘Triangles’ is one of the most important chapters of Class 10. So, students must solve every exercise of the given chapter, and Extramarks has provided every solution to each question of Chapter 6- Triangles. The first exercise 6.1 has a total number of 3 questions which can be accessible from NCERT Solutions For Class 10 Maths Chapter 6 Exercise 6.1. The second exercise 6.2 has a total number of 10 questions, and the NCERT Solutions For Class 10 Maths Chapter 6 Exercise 6.2 help solve those questions. Further, the third exercise 6.3 has a total number of 16 questions that can be accessed from NCERT Solutions for Class 10 Maths Chapter 6 Exercise 6.3, the fourth exercise 6.4 has a total number of 9 questions that can be accessed from NCERT Solutions for Class 10 Maths Chapter 6 Exercise 6.4, the fifth exercise 6.5 has a total number of 17 questions that can be accessed from NCERT Solutions for Class 10 Maths Chapter 6 Exercise 6.5, and lastly, the sixth exercise has a total number of 10 questions that can be accessible from NCERT Solutions For Class 10 Maths Chapter 6 Exercise 6.5.So, students must solve every exercise of the given chapter, and Extramarks has provided every solution to each question of Chapter 6- Triangles. Students are advised to solve each and every question inthe chapter so that they will not face any issues while solving questions in the examination hall. Students must go through each exercise and should focus on NCERT Solutions For Class 10 Maths Chapter 6 Exercise 6.2.

### Key Features for Exercise 6.2 Class 10 NCERT Solutions

The theorems explained in the NCERT Solutions For Class 10 Maths Chapter 6 Exercise 6.2 must be thoroughly understood. Therefore, it is suggested that students read the logical explanations of the theorems’ proofs and apply them using clear illustrations.

If the students are able to remember and comprehend the theorems’ applications, they will be able to successfully comprehend the NCERT Solutions For Class 10 Maths Chapter 6 Exercise 6.2. Students might also think about creating a theorem chart that they can occasionally refer to. Using common items is a fun and simple way to study theorems. Students can do the same by asking their parents and teachers for assistance and always getting access to NCERT Solutions For Class 10 Maths Chapter 6 Exercise 6.2.

**Q.1** In the following figure, (i) and (ii), DE || BC. Find EC in (i) and AD in (ii).

**Ans.**

(i)

$\begin{array}{l}\text{It is given that}\mathrm{DE}\parallel \mathrm{BC}.\\ \text{By using Basic Proportionality Theorem, we get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{\mathrm{AD}}{\mathrm{DB}}=\frac{\mathrm{AE}}{\text{\hspace{0.17em}}\mathrm{EC}}\\ \Rightarrow \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{1.5}{3}=\frac{1}{\mathrm{EC}}\\ \Rightarrow \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{EC}=\frac{3\times 1}{1.5}=2\\ \therefore \mathrm{EC}=2\text{cm}\end{array}$

(ii)

$\begin{array}{l}\text{It is given that}\mathrm{DE}\text{\hspace{0.17em}\hspace{0.17em}}\left|\right|\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{BC}.\\ \text{By using Basic Proportionality Theorem, we get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{\mathrm{AD}}{\mathrm{DB}}=\frac{\mathrm{AE}}{\text{\hspace{0.17em}}\mathrm{EC}}\\ \Rightarrow \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{\mathrm{AD}}{7.2}=\frac{1.8}{5.4}=\frac{1}{3}\\ \Rightarrow \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{AD}=\frac{7.2}{3}=2.4\\ \therefore \mathrm{AD}=2.4\text{cm}\end{array}$

**Q.2 **

$\begin{array}{l}\text{E and F are points on the sides PQ and PR respectively of a}\mathrm{\Delta}\text{PQR. For each of the following}\\ \text{cases, state whether EF}\parallel \text{QR:}\\ \text{(i)}\mathrm{PE}=3.9\text{cm,}\mathrm{EQ}=3\text{cm,}\mathrm{PF}=3.6\text{cm and}\mathrm{FR}=2.4\text{cm}\\ \text{(ii)}\mathrm{PE}=4\text{cm,}\mathrm{QE}=4.5\text{cm,}\mathrm{PF}=8\text{cm and}\mathrm{RF}=9\text{cm}\\ \text{(iii)}\mathrm{PQ}=1.28\text{cm,}\mathrm{PR}=2.56\text{cm,}\mathrm{PE}=0.18\text{cm and}\mathrm{PF}=0.36\text{cm}\end{array}$

**Ans.**

(i)

As per given information, we have the following triangle.

$\begin{array}{l}\text{In the above triangle},\\ \frac{\mathrm{PE}}{\mathrm{EQ}}=\frac{3.9}{3}=1.3\\ \text{and}\\ \frac{\mathrm{PF}}{\mathrm{FR}}=\frac{3.6}{2.4}=\frac{3}{2}=1.5\\ \text{Hence,}\frac{\mathrm{PE}}{\mathrm{EQ}}\ne \frac{\mathrm{PF}}{\mathrm{FR}}\\ \text{i.e., the line}\mathrm{EF}\text{does not divide}\mathrm{PQ}\text{and}\mathrm{PR}\text{in}\\ \text{the same ratio and so}\mathrm{EF}\text{is not parallel to}\mathrm{QR}.\end{array}$

(ii)

As per given information, we have the following triangle.

$\begin{array}{l}\text{In the above triangle},\\ \frac{\mathrm{PE}}{\mathrm{EQ}}=\frac{4}{4.5}=\frac{40}{45}=\frac{8}{9}\\ \text{and}\\ \frac{\mathrm{PF}}{\mathrm{FR}}=\frac{8}{9}\\ \text{Hence,}\frac{\mathrm{PE}}{\mathrm{EQ}}=\frac{\mathrm{PF}}{\mathrm{FR}}\\ \text{i.e., the line}\mathrm{EF}\text{divides}\mathrm{PQ}\text{and}\mathrm{PR}\text{in}\\ \text{the same ratio and so}\mathrm{EF}\text{is parallel to}\mathrm{QR}.\end{array}$

(iii)

As per given information, we have the following triangle.

$\begin{array}{l}\left(\mathrm{iv}\right)\\ \text{In the above triangle},\\ \frac{\mathrm{PE}}{\mathrm{EQ}}=\frac{0.18}{1.1}=\frac{18}{110}=\frac{9}{55}\\ \text{and}\\ \frac{\mathrm{PF}}{\mathrm{FR}}=\frac{0.36}{2.2}=\frac{36}{220}=\frac{9}{55}\\ \text{Hence,}\frac{\mathrm{PE}}{\mathrm{EQ}}=\frac{\mathrm{PF}}{\mathrm{FR}}\\ \text{i.e., the line}\mathrm{EF}\text{divides}\mathrm{PQ}\text{and}\mathrm{PR}\text{in}\\ \text{the same ratio and so}\mathrm{EF}\text{is parallel to}\mathrm{QR}.\end{array}$

**Q.3 **

$\begin{array}{l}\text{In the following figure, if}\mathrm{LM}\parallel \mathrm{CB}\text{and}\mathrm{LN}\parallel \mathrm{CD},\text{prove that}\\ \frac{\mathrm{AM}}{\mathrm{AB}}=\frac{\mathrm{AN}}{\mathrm{AD}}.\end{array}$

**Ans.**

$\begin{array}{l}\text{In the given figure,}\mathrm{LM}\parallel \mathrm{CB}\\ \text{So, using basic proportionality theorem in}\mathrm{\Delta}\text{ABC, we get}\\ \frac{\mathrm{AM}}{\mathrm{AB}}=\frac{\mathrm{AL}}{\mathrm{AC}}\text{}...\text{(1)}\\ \text{Also, in the given figure,}\mathrm{LN}\parallel \mathrm{CD}\\ \text{So, using basic proportionality theorem in}\mathrm{\Delta}\text{ACD, we get}\\ \frac{\mathrm{AN}}{\mathrm{AD}}=\frac{\mathrm{AL}}{\mathrm{AC}}\text{}...\text{(2)}\\ \text{From (1) and (2), we get}\\ \frac{\mathrm{AM}}{\mathrm{AB}}=\frac{\mathrm{AN}}{\mathrm{AD}}.\end{array}$

**Q.4 **

$\begin{array}{l}\text{In the following figure,}\mathrm{DE}\parallel \mathrm{AC}\text{and}\mathrm{DF}\parallel \mathrm{AE}.\text{}\\ \text{Prove that}\frac{\mathrm{BF}}{\mathrm{FE}}=\frac{\mathrm{BE}}{\mathrm{EC}}.\end{array}$

**Ans.**

$\begin{array}{l}\text{In}\mathrm{\Delta}\text{ABC,}\mathrm{DE}\parallel \mathrm{AC}\\ \text{So, using basic proportionality theorem, we get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{\mathrm{BD}}{\mathrm{DA}}=\frac{\mathrm{BE}}{\mathrm{EC}}\text{}...\text{(1)}\\ \text{In}\mathrm{\Delta}\text{BAE,}\mathrm{DF}\parallel \mathrm{AE}\\ \text{So, using basic proportionality theorem, we get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{\mathrm{BD}}{\mathrm{DA}}=\frac{\mathrm{BF}}{\mathrm{FE}}\text{}...\text{(2)}\\ \text{From (1) and (2), we get}\\ \frac{\mathrm{BE}}{\mathrm{EC}}=\frac{\mathrm{BF}}{\mathrm{FE}}\end{array}$

**Q.5 **

**Ans.**

$\begin{array}{l}\text{In}\mathrm{\Delta}\text{POQ,}\mathrm{DE}\parallel \mathrm{OQ}\\ \text{So, using basic proportionality theorem, we get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{\mathrm{PE}}{\mathrm{EQ}}=\frac{\mathrm{PD}}{\mathrm{DO}}\text{}...\text{(1)}\\ \text{In}\mathrm{\Delta}\text{POR,}\mathrm{DF}\parallel \mathrm{OR}\\ \text{So, using basic proportionality theorem, we get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{\mathrm{PF}}{\mathrm{FR}}=\frac{\mathrm{PD}}{\mathrm{DO}}\text{}...\text{(2)}\\ \text{From (1) and (2), we get}\\ \frac{\mathrm{PE}}{\mathrm{EQ}}=\frac{\mathrm{PF}}{\mathrm{FR}}\\ \text{Hence, by converse of basic proportionality theorem,}\mathrm{EF}\parallel \mathrm{QR}.\end{array}$

**Q.6** In the following figure, A, B and C are points on OP, OQ and OR respectively such that AB ∥ PQ and AC ∥ PR.Show that BC ∥ QR.

**Ans.**

$\begin{array}{l}\text{In \Delta POQ, AB\hspace{0.33em}ll\hspace{0.33em}PQ (Given)}\\ \text{Therefore, by Basic Proportionality Theorem,}\\ \frac{\text{AP}}{\text{AO}}\text{=}\frac{\text{OB}}{\text{BQ}}\text{}...\text{(i)}\\ \text{In \Delta POR, AC\hspace{0.33em}ll\hspace{0.33em}PR (Given)}\\ \text{Therefore, by Basic Proportionality Theorem,}\\ \frac{\text{AP}}{\text{AO}}\text{=}\frac{\text{OC}}{\text{CR}}\text{}...\text{(ii)}\\ \text{From (i) and (ii),}\\ \frac{\text{OB}}{\text{BQ}}\text{=}\frac{\text{OC}}{\text{CR}}\text{}\\ \text{Hence, by using Converse of Basic Proportionality Theorem}\\ \text{in \Delta OQR, we find that BC\hspace{0.33em}ll\hspace{0.33em}QR.}\end{array}$

**Q.7 ** Using Basic Proportionality Theorem, prove that a line drawn through the mid-point of one side of a triangle parallel to another side bisects the third side. (Recall that you have proved it in Class IX).

**Ans.**

We consider ΔABC drawn below in which D is the mid-point of side AB and DE is the line segment drawn parallel to BC.

$\begin{array}{l}\text{By using Basic Proportionality Theorem, we get}\\ \frac{\text{AD}}{\text{DB}}\text{\hspace{0.33em}=\hspace{0.33em}}\frac{\text{AE}}{\text{EC}}\text{}...\text{(1)}\\ \text{D is the mid-point of AB. Therefore, AD=DB}\\ \text{From (1), we have}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{\text{AD}}{\text{DB}}\text{\hspace{0.33em}=\hspace{0.33em}}\frac{\text{AE}}{\text{EC}}\text{}\\ \Rightarrow \frac{\text{AD}}{\text{AD}}\text{\hspace{0.33em}=\hspace{0.33em}}\frac{\text{AE}}{\text{EC}}\text{}\\ \Rightarrow \text{1\hspace{0.33em}=\hspace{0.33em}}\frac{\text{AE}}{\text{EC}}\\ \mathrm{\Rightarrow}\text{EC\hspace{0.33em}=\hspace{0.33em}AE}\\ \text{Therefore, E is the mid-point of AC.}\end{array}$

**Q.8** Using converse of Basic Proportionality Theorem, prove that the line joining the mid-points of any two sides of a triangle is parallel to the third side.

(Recall that you have done it in Class IX)

**Ans.**

We consider ΔABC drawn below in which DE is the line segment joining the mid-points D and E of sides AB and AC respectively.

$\begin{array}{l}\text{We have}\\ \mathrm{AD}=\mathrm{DB}\Rightarrow \frac{\mathrm{AD}}{\mathrm{DB}}=1\\ \text{and}\\ \mathrm{AE}=\mathrm{EC}\Rightarrow \frac{\mathrm{AE}}{\mathrm{EC}}=1\\ \text{Therefore,}\\ \frac{\mathrm{AD}}{\mathrm{DB}}=\frac{\mathrm{AE}}{\mathrm{EC}}\\ \text{Hence, by converse of Basic Proportionality Theorem,}\mathrm{DE}\parallel \mathrm{BC}.\end{array}$

**Q.9 **

$\begin{array}{l}\mathrm{ABCD}\text{is a trapezium in which}\mathrm{AB}\parallel \mathrm{DC}\text{and its diagonals intersect each other at the point}\mathrm{O}.\\ \text{Show that}\frac{\mathrm{AO}}{\mathrm{BO}}=\frac{\mathrm{CO}}{\mathrm{DO}}.\end{array}$

**Ans.**

$\begin{array}{l}\text{Given}:\\ \text{ABCD is a trapezium in which AB is parallel to CD and diagonals intersect each other at O}.\end{array}$

$\begin{array}{l}\text{To prove}:\\ \frac{\mathrm{AO}}{\mathrm{BO}}=\frac{\mathrm{CO}}{\mathrm{DO}}\\ \text{Construction}:\\ \text{Draw}\mathrm{OF}\parallel \mathrm{AB}\text{which meets AD in F.}\\ \text{In}\mathrm{\Delta ABD}\text{,}\mathrm{OF}\parallel \mathrm{AB}\\ \text{So, by converse of Basic\hspace{0.17em}\hspace{0.17em}Proportionality Theorem,}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{\mathrm{DO}}{\mathrm{OB}}=\frac{\mathrm{DF}}{\mathrm{FA}}\text{}...\text{(i)}\\ \text{Now in}\mathrm{\Delta ADC},\text{}\mathrm{OF}\parallel \mathrm{CD}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}[\mathrm{S}\text{ince}\mathrm{AB}\parallel \mathrm{CD}]\\ \therefore \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{\mathrm{DF}}{\mathrm{FA}}=\frac{\mathrm{CO}}{\mathrm{OA}}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\text{(ii)}\\ \text{Comparing equation (i) and (ii), we get}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{\mathrm{DO}}{\mathrm{OB}}=\frac{\mathrm{CO}}{\mathrm{OA}}\\ \Rightarrow \text{}\frac{\mathrm{OA}}{\mathrm{OB}}=\frac{\mathrm{CO}}{\mathrm{DO}}\end{array}$

**Q.10 ** ** **

$\begin{array}{l}\text{The diagonals of a quadrilateral ABCD intersect each other at the point O such that}\\ \frac{\text{AO}}{\text{BO}}\text{\hspace{0.33em}=\hspace{0.33em}}\frac{\text{CO}}{\text{DO}}\text{. Show that\hspace{0.33em}ABCD is a trapezium.}\end{array}$

**Ans.**

$\begin{array}{l}\text{Given}:\\ \text{Diagonals of a quadrilateral ABCD intersect each other at point O such that}\frac{\mathrm{AO}}{\mathrm{BO}}=\frac{\mathrm{CO}}{\mathrm{DO}}.\end{array}$

$\begin{array}{l}\text{To prove}:\\ \mathrm{AB}\parallel \text{CD}\\ \text{Construction}:\\ \text{Draw}\mathrm{OF}\parallel \mathrm{AB}\text{which meets AD in F.}\\ \text{In}\mathrm{\Delta ABD}\text{,}\mathrm{OF}\parallel \mathrm{AB}\\ \text{So, by converse of Basic\hspace{0.17em}\hspace{0.17em}Proportionality Theorem,}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{\mathrm{DO}}{\mathrm{BO}}=\frac{\mathrm{DF}}{\mathrm{FA}}\text{}...\text{(i)}\\ \mathrm{Given}\text{}\mathrm{that}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{\mathrm{AO}}{\mathrm{BO}}=\frac{\mathrm{CO}}{\mathrm{DO}}\\ \Rightarrow \frac{\mathrm{DO}}{\mathrm{BO}}=\frac{\mathrm{CO}}{\mathrm{AO}}\\ \Rightarrow \frac{\mathrm{DF}}{\mathrm{FA}}=\frac{\mathrm{CO}}{\mathrm{AO}}\text{[From (i)]}\\ \Rightarrow \text{OF}\parallel \text{DC [By converse of Basic Proportionality Theorem]}\\ \text{Also, OF}\parallel \mathrm{AB}\\ \mathrm{Therefore},\text{}\mathrm{AB}\parallel \text{DC}\\ \mathrm{Hence},\text{ABCD is a trapezium.}\end{array}$

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