# NCERT Solutions for Class 9 Maths Chapter 7 Triangles Exercise 7.3

A student’s life begins to turn when they reach CBSE Class 10. It affects their higher education options and careers. However, a student’s success is not solely based on how well they perform during that particular academic year. Class 9 and Class 10 are similar to two sides of the same coin, indicating the significance of CBSE Class 9 as a life-changing experience for students. Students must pass Class 9 to sit for the Class 10 board exams. Additionally, it lays the groundwork for subsequent classes.

It will be simpler for a student to achieve high marks in the exams for the higher classes if they have a thorough understanding of the Class 9 subjects. Math is such a subject that demands a lot of practice. The papers provide students with practice while also covering the concepts of important topics in Class 9 subjects.

Among the most crucial subjects is mathematics. The elements of mathematics include numbers, shapes, data, measurements, and logical operations. Every area of life, including medicine, engineering, finance, the natural sciences, economics, etc., has a significant application for it. The world of mathematics is all around us.

There are numerous real-world applications for the mathematical ideas, theories, and formulas students learn in math textbooks. Students must learn the concepts and formulas in order to solve a variety of issues. It is crucial to understand this subject to comprehend its many applications and significance.

Math fundamentals like place value, addition, subtraction, multiplication, and counting are taught at the beginning level. Students are taught more complex concepts as they advance in grade level, including ratios, proportions, fractions, algebra, geometry, trigonometry, measurement, etc. The more advanced topics in the higher secondary school curriculum include integration and differentiation.

Mathematics is a discipline of rationality; it is the science of structural relations and order, which has its roots in primitive counting, measuring, and object-shape explanation techniques. Additionally, it deals with logical reasoning and quantitative calculation. Thus, the word “Mathematics” means “to study, learn, or acquire knowledge.” Mathematical theories assist us in comprehending and resolving a variety of issues in both academic and practical contexts.

Mathematics progressed, its subject became more idealized and abstracted. Numerous Mathematicians have studied the topic for many years in various civilizations across the globe. Archimedes (287–212 BC) is regarded as the founder of Mathematics and lived during the BC century. He developed formulas for figuring out a solid’s volume and surface area. Aryabhatt, who was born in 476 CE, is applauded as the father of Indian Mathematics. Mathematics has been a crucial complement to technology and the physical sciences since the 17th century. Recently, it has been assumed that it functions similarly in the quantitative measures of the life sciences.

In the sixth century BC, the Pythagoreans established the study of Mathematics as a “demonstrative discipline.” The Greek word “mathema,” which means “the subject of instruction,” is where the word “Mathematics” originated. Axioms, theorems, proofs, and postulates were developed by another mathematician named Euclid. These concepts are still used extensively in modern Mathematics.

The early study of Mathematics history uses a differential approach to define Mathematics history in every region of the world. Many concepts have different theories that various Mathematicians have developed, and those are now applied in modern Mathematics.

The BODMAS rule is the one that is most frequently applied in mathematics. According to this rule, the order of operations and the brackets determine how the arithmetic operations are carried out. Students can quickly comprehend this reasoning by using the complete form of BODMAS. Brackets Orders, Division, Multiplication, Addition, and Subtraction, or BODMAS Students can remember the order of operations in calculations with the help of the acronym BODMAS rule. The various Mathematical operations on numbers are known as “operations.”

The definition of “order” in BODMAS can sometimes be a little unclear. Order is just another way of saying square roots or square numbers. BODMAS is another name for BODMAS. When a number is multiplied by itself, it is indicated by a small number called an index, represented by the letter I in that acronym (that it is a square number).

Using the BODMAS rule entails solving any calculations enclosed in brackets first, then determining the square or square root of any numbers, and so forth, with any subtraction calculations coming last. However, it’s crucial to remember that multiplication and division should be finished in whichever order comes first, from left to right. Similar rules apply to addition and subtraction.

The area of mathematics known as geometry concerns the dimensions, sizes, shapes, and angles of a wide range of everyday objects. The words “geometry” and “metron,” both of which mean “measurement,” are derived from Ancient Greek. Both two-dimensional and three-dimensional shapes exist in Euclidean geometry.

Plain shapes in plane geometry include 2-dimensional shapes like triangles, squares, rectangles, and circles. Solids in solid geometry are also three-dimensional shapes like a cube, cuboids, cones, etc. According to coordinate geometry, the fundamental building blocks of geometry are points, lines, and planes. Geometry’s various shapes make it easier for us to comprehend the shapes we encounter daily. We can determine shapes’ area, perimeter, and volume using geometrical concepts.

A student’s life is turning when they reach CBSE Class 10. It affects their higher education options and career. However, a student’s success is not solely based on how well they perform during that particular academic year. Class 9 and Class 10 are similar to two sides of the same coin, indicating the significance of CBSE Class 9 as a life-changing experience for students. Students must pass Class 9 to sit for the Class 10 board exams. Additionally, it lays the groundwork for subsequent classes.

It will be simpler for a student to achieve high marks in the exams for the higher classes if they have a thorough understanding of the Class 9 subjects. A student’s choice of the stream for Class 10 becomes more difficult when they are in Class 9. Strong fundamental knowledge and mental aptitude are essential for passing the entrance exams. Building the concepts starting in Class 9 is necessary for this kind of understanding.

Most students don’t start studying for entrance examinations until Class 11; instead, they skip Class 9 and concentrate only on getting high marks on the board exam. It frequently arrives quite late for exams with such a high bar.

Therefore, one must start developing aptitude as soon as possible. Whether a student is preparing for the JEE-Advanced, AIIMS, JEE-Mains, NEET, or another exam, the earlier they start is preferable.

Additionally, the way science and math are taught in Classes 9 and 10 tends to emphasize cramming for exams rather than conceptual learning. The students become confused as a result of this. As a result, a student promoted to Class 11 feels that the difficulty level in the subjects has increased significantly from Classes 9 and 10. Building on the concepts from class 9 becomes even more crucial.

The fundamental Newton’s laws of motion and the three equations of motion are introduced to students in Class 9. It serves as the cornerstone of introductory kinematics. It necessitates a thorough comprehension of motion and associated ideas. Similarly, finding lengths and areas is a primary goal of the Class 9 course Coordinate Geometry, which also deals with the 2-dimensional rectangular coordinate system. This concept is expanded to three dimensions while preparing for entrance exams, and it has many uses in topics of other subjects as well, such as physics.

Even though they are not directly related to the entrance exams, some topics are applied to create the foundation for other, more complex topics. To adequately prepare for the entrance exams, it is therefore important to be familiar with them. Only two years are given to students who begin their entrance exam preparations in class eleven to complete the extensive course requirements. Those who begin with Class 9 can finish it at a more leisurely pace. They have a particular advantage over their classmates in the four years they have had to develop and comprehend the concepts. The early bird does, after all, get the worm.

The early starters will have plenty of time for in-depth revision and practice during the final year, when most students will be busy finishing their courses. They won’t also be in a rush to finish the syllabus because they have already mastered the necessary ideas.

Additionally, the students can efficiently plan their studies because they are already familiar with the subjects. The student’s sense of security will increase thanks to early planning and effective management. Thus, developing concepts in Class 9 is critical for a student’s knowledge and performance. So, begin early, develop student’s concepts, and get a head start on the future.

Extramarks is one of the best learning platforms for all students. Students know that Mathematics is one of the subjects that requires a lot of practice. The NCERT Solutions For Class 9 Maths Chapter 7 Exercise 7.3 will help students to understand all the important topics. Sometimes students get confused while doing practice. They can take help from the NCERT Solutions For Class 9 Maths Chapter 7 Exercise 7.3. Students can also download the NCERT Solutions For Class 9 Maths Chapter 7 Exercise 7.3 offline from the Extramarks website. The highly experienced Mathematics teachers created the NCERT Solutions For Class 9 Maths Chapter 7 Exercise 7.3 to help the students.

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**NCERT Solutions for Class 9 Maths Chapter 7 Triangles (Ex 7.3) Exercise 7.3**

Chapter 7 of the NCERT for Class 9 Mathematics is crucial because it teaches students about triangles. The various kinds of triangles and their properties are covered in this chapter for the benefit of the students. Many significant problems, equations, and theorems are also included in this chapter.

The Extramarks website gives students access to the NCERT Solutions For Class 9 Maths Chapter 7 Exercise 7.3. At Extramarks, experienced teachers discussed everything students need to know about this chapter in NCERT Solutions For Class 9 Maths Chapter 7 Exercise 7.3. The NCERT Solutions For Class 9 Maths Chapter 7 Exercise 7.3 are available in PDF format. Moreover, from the Extramarks website, students can quickly download NCERT Solutions For Class 9 Maths Chapter 7 Exercise 7.3.

One of the most fundamental two-dimensional geometric shapes is a triangle, which is created by the intersection of three lines. Triangle inequalities are discussed in conjunction with rules of congruence, the congruence of triangles, and other properties in the NCERT Solutions For Class 9 Maths Chapter 7 Exercise 7.3, Triangles. Beginning with a brief introduction to triangles, their types, and their properties, this chapter ensures students understand these shapes. NCERT Solutions For Class 9 Maths Chapter 7 Exercise 7.3, covers significant formulas, problems, and theorems based on congruent triangles and triangle inequalities.

**Access NCERT Solutions for Math Class 9 Chapter 7 – Triangles**

The Class 9 Mathematics Chapter 7 Triangle contains five exercises explaining the topics. The student’s knowledge of triangles and their characteristics will improve. Understanding various complex concepts, such as congruency in triangles, necessitates using the explicit geometric shape known as the triangle. The sides and angles of two triangles must match to be considered congruent. To effectively use triangles, it is essential to fully comprehend their fundamental characteristics. It will also provide a solid foundation for future geometry studies to have a solid understanding of triangles. Students can use NCERT Solutions For Class 9 Maths Chapter 7 Exercise 7.3, to strengthen their understanding of triangles and the questions in each section.

The NCERT Solutions For Class 9 Maths Chapter 7 Exercise 7.3 make things easy to understand the Chapter easily. Students can also check their solution with NCERT Solutions For Class 9 Maths Chapter 7 Exercise 7.3. NCERT Solutions For Class 9 Maths Chapter 7 Exercise 7.3 are easily downloaded from the Extramarks website. Students can also get the NCERT Solutions For Class 9 Maths Chapter 7 Exercise 7.3 in PDF format.

**Exercise (7.1)**

The “Congruence of Triangles” is the topic of Exercise 7.1 in Chapter 7, Class 9. Although students hardly see applications for this topic, they must be perplexed about why it was included in the curriculum. Two triangles are rarely congruent in reality, as students might imagine. However, studying this subject becomes essential because it is utilized to create substantial buildings or architectural designs. The Mathematician discovered the Triangle to be the most stable shape, and congruence is required to produce even surfaces. The geometric art, carpet patterns, stepping stone patterns, architectural designs, etc., must have caught students’ attention for having congruent triangles.

Eight questions based on the idea of triangle congruence and various standards for proving triangle congruence make up the first exercise of Chapter 7, Triangles in CBSE Maths Class 9. Although students might question the need to demonstrate two triangles are congruent given how rarely this occurs in real life, the topic of Triangles in Mathematical Chapter 7 is quite significant practically.

Exercise 7.1 consists of eight questions, all focusing on using various rules to demonstrate the congruence of triangles. Students should remember each step while solving these problems because this will be the first time they encounter them in Class 9 Mathematics.

This first exercise from Chapter 7, Triangles, applies the foundation for the entire chapter. Before attempting Chapter 7 Maths NCERT Solutions, students must understand the chapter’s significance and pay close attention to each theorem and axiom of congruence criteria.

**Exercise (7.2)**

Triangle properties and associated theorems are covered in Chapter 7’s Exercise 7.2. Since this topic focuses more on proving than solving, most students find it boring. However, if students can relate the Triangle’s properties to the examples, they will become interested in them.

Students reach the NCERT Solutions For Class 9 Maths Chapter 7 Exercise 7.3, first complete the previous exercises and understand the concepts. Students can also download the NCERT Solutions from the Extramarks for other exercises.

**Exercise (7.3)**

A few more requirements for the congruence of triangles are covered in the third exercise of Chapter 7 in Mathematics. The main emphasis of this exercise will be on the SSS and RHS congruence rules. There are only five proof-based questions in this exercise. NCERT Solutions For Class 9 Maths Chapter 7 Exercise 7.3 provided solutions to all questions in PDF format with detailed explanations to aid students in their studies. NCERT Solutions For Class 9 Maths Chapter 7 Exercise 7.3, can be downloaded by students from the Extramarks website.

Extramarks is one of the best learning websites for students. NCERT Solutions For Class 9 Maths Chapter 7 Exercise 7.3 help students to understand the chapter in-depth. Students can also take help from the NCERT Solutions For Class 9 Maths Chapter 7 Exercise 7.3 when they stop at some point in solving the question. NCERT Solutions For Class 9 Maths Chapter 7 Exercise 7.3 also helps students to know how to solve the questions step-wise and get the best marks in exams.

**Exercise (7.4)**

Students learn about the newest concept of triangles in Exercise 7.4, which deals with the triangle’s angles and side congruence. It enables them to comprehend how the two triangles are similar. Through Exercise 7.4, students receive the assistance they need to understand the concept fully. The differences between the two triangles will teach them more about how they differ from one another. Students can find chapter-by-chapter guidance in the NCERT Solutions For Class 9 Maths Chapter 7 Exercise 7.3. The NCERT Solutions For Class 9 Maths Chapter 7 Exercise 7.3 study material is available at any time for a thorough understanding.

**Exercise (7.5)**

Exercise 7.5 is not required, so the exam will not include any questions from it. Now that this exercise is available in the NCERT book, students must wonder why. The addition is being made to improve the students’ capacity for thought, is the answer. The chapters in the NCERT Maths textbook are not only included for exam-passing purposes. The main objective is to produce outstanding students capable of solving problems in the real world and possessing theoretical knowledge. Therefore, this exercise’s main purpose is to test students’ understanding of the material they have already learned in chapter 7, Triangles.

**NCERT Solutions For Class 9 Maths Chapter 7 Exercise 7.3**

The NCERT Solutions For Class 9 Maths Chapter 7 Exercise 7.3 are a great way to review and strengthen concepts from the new curriculum. The NCERT Solutions For Class 9 Maths Chapter 7 Exercise 7.3 can help support students in understanding the subject and quickly grasping a specific topic or exercise.

Make sure the students comprehend the fundamentals of calculating angles in a triangle. Students need to understand the idea of angles adding up to 180 degrees and the rule that the longest side of a triangle is always on the other side of the largest angle. Students should be able to approach the exercise confidently once they have that idea down. NCERT Solutions For Class 9 Maths Chapter 7 Exercise 7.3 can help students who are stuck on any of the questions in Class 9 Ex 7.3.

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**NCERT Solutions Class 9 Maths Chapter 7 Exercise 7.3**

Students can download NCERT Solutions For Class 9 Maths Chapter 7 Exercise 7.3 in PDF format. These solutions have been prepared by expert mathematics teachers and cover all of the topics and exercises in the NCERT textbook. So, whether students are having difficulty with questions or anything else, NCERT Solutions For Class 9 Maths Chapter 7 Exercise 7.3 will help them.

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**Class 9 Maths Chapter 7 Triangles Exercise 7.3- Weightage Marks**

Mathematics Class 9 exams weigh 22 marks for triangles, which are part of the Geometry unit. NCERT Solutions For Class 9 Maths Chapter 7 Exercise 7.3 help students get the desired mathematics exam marks. Moreover, NCERT Solutions For Class 9 Maths Chapter 7 Exercise 7.3 are best for revising Chapter 7. Even students match their practice solutions with NCERT Solutions For Class 9 Maths Chapter 7 Exercise 7.3 to check the right answer.

**Benefits of Class 9 Maths Chapter 7 Exercise 7.3 NCERT Solutions**

Numerous advantages come from studying NCERT Solutions For Class 9 Maths Chapter 7 Exercise 7.3. Students will better understand the subjects, which is its main benefit. All of the questions from Exercise 7.3 Class 9 Maths are answered with the NCERT Solutions For Class 9 Maths Chapter 7 Exercise 7.3, along with thorough explanations to help students better understand the material and increase their practice in this area. Before students can solve a problem, they must understand how to solve it. This is why students must study NCERT Solutions For Class 9 Maths Chapter 7 Exercise 7.3, before taking any exams or test papers.

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**NCERT Solutions for Class 9**

All the questions in the Class 9 NCERT textbook have solutions in the NCERT Solutions for Mathematics in Class 9. Through the links provided below on this page, students can download PDFs of the solutions to these problems, broken down by chapter. All of the topics covered in the NCERT textbook for Class 9 are covered in these NCERT Solutions, including the number system, coordinate geometry, polynomials, Euclidean geometry, quadrilaterals, triangles, circles, constructions, surface areas and volumes, statistics, probability, etc.

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**CBSE Study Materials for Class 9 CBSE Study Materials**

As students are introduced to new and challenging ideas and concepts that will be further explored in higher classes, CBSE Class 9 is regarded as a crucial time in their academic careers. In order to succeed in the higher standards, it is, therefore, preferable to clear the fundamentals in class nine itself. CBSE Class 9 Study Material was created to prepare for the upcoming academic year.

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**Q.1 **

$\begin{array}{l}\mathrm{\Delta ABC}\mathrm{and}\mathrm{\Delta DBC}\mathrm{are}\mathrm{two}\mathrm{isosceles}\mathrm{triangle}\mathrm{son}\mathrm{the}\mathrm{same}\\ \mathrm{base}\mathrm{BC}\mathrm{and}\mathrm{vertices}\mathrm{A}\mathrm{and}\mathrm{D}\mathrm{are}\mathrm{on}\mathrm{the}\mathrm{same}\mathrm{side}\mathrm{of}\mathrm{BC}\\ (\mathrm{see}\mathrm{the}\mathrm{figure}\mathrm{below}).\mathrm{If}\mathrm{AD}\mathrm{is}\mathrm{extended}\mathrm{to}\mathrm{intersect}\mathrm{BC}\mathrm{at}\mathrm{P},\mathrm{show}\\ \text{that}\end{array}$

$\begin{array}{l}\left(\mathrm{i}\right)\text{\hspace{0.33em}}\mathrm{\Delta ABD}\cong \mathrm{\Delta ACD}\\ \left(\mathrm{ii}\right)\text{\hspace{0.33em}}\mathrm{\Delta ABP}\cong \mathrm{\Delta ACP}\\ \left(\mathrm{iii}\right)\text{\hspace{0.33em}}\mathrm{AP}\mathrm{bisects}\angle \mathrm{A}\mathrm{as}\mathrm{well}\mathrm{as}\angle \mathrm{D}.\\ \left(\mathrm{iv}\right)\text{\hspace{0.33em}}\mathrm{AP}\mathrm{is}\mathrm{the}\mathrm{perpendicular}\mathrm{bisector}\mathrm{of}\mathrm{BC}.\end{array}$

**Ans**

$\begin{array}{l}\hspace{0.17em}\hspace{0.17em}\mathrm{Given}:\mathrm{\Delta ABC}\mathrm{and}\mathrm{\Delta DBC}\mathrm{are}\mathrm{two}\mathrm{isosceles}\mathrm{triangles}\mathrm{on}\mathrm{the}\mathrm{same}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{base}\mathrm{BC}\mathrm{}\mathrm{and}\mathrm{vertices}\mathrm{A}\mathrm{and}\mathrm{D}\mathrm{are}\mathrm{on}\mathrm{the}\mathrm{same}\mathrm{side}\mathrm{of}\mathrm{BC}.\\ \mathrm{To}\mathrm{prove}:\mathrm{\hspace{0.17em}}\left(\mathrm{i}\right)\mathrm{\Delta ABD}\cong \mathrm{\Delta ACD}\\ \mathrm{\hspace{0.17em}}\left(\mathrm{ii}\right)\mathrm{\Delta ABP}\cong \mathrm{\Delta ACP}\\ \mathrm{\hspace{0.17em}}\left(\mathrm{iii}\right)\mathrm{AP}\mathrm{bisects}\angle \mathrm{A}\mathrm{as}\mathrm{well}\mathrm{as}\angle \mathrm{D}.\\ \\ \mathrm{\hspace{0.17em}}\left(\mathrm{iv}\right)\mathrm{AP}\mathrm{is}\mathrm{the}\mathrm{perpendicular}\mathrm{bisector}\mathrm{of}\mathrm{BC}.\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{Proof}:\left(\mathrm{i}\right)\mathrm{In}\mathrm{}\mathrm{\Delta ABD}\hspace{0.17em}\hspace{0.17em}\mathrm{and}\mathrm{\Delta ACD}\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{AB}=\mathrm{AC}\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{\hspace{0.17em}}\left[\mathrm{Given}\right]\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{DB}=\mathrm{DC}\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{\hspace{0.17em}}\left[\mathrm{Given}\right]\\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}\mathrm{AD}=\mathrm{AD}\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{\hspace{0.17em}}\left[\mathrm{Common}\right]\\ \therefore \mathrm{\Delta ABD}\mathrm{\hspace{0.17em}}\cong \mathrm{\Delta ACD}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}[\mathrm{By}\mathrm{S}.\mathrm{S}.\mathrm{S}\mathrm{.}]\\ \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{BAD}=\angle \mathrm{CAD}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{\hspace{0.17em}}[\mathrm{By}\mathrm{C}.\mathrm{P}.\mathrm{C}.\mathrm{T}\mathrm{.}]\end{array}$

$\begin{array}{l}\left(\mathrm{ii}\right)\text{\hspace{0.33em}}\mathrm{In}\text{}\mathrm{\Delta ABP}\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{and\Delta ACP}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{AB}=\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{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}}\angle \mathrm{BAP}=\angle \mathrm{CAP}\text{\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}}\mathrm{AP}=\mathrm{AP}\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]\\ \therefore \mathrm{\Delta ABP}\text{\hspace{0.17em}}\cong \mathrm{\Delta ACP}\text{\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}}\angle \mathrm{BAP}=\angle \mathrm{CAP}\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}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{BP}=\mathrm{CP}\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]\\ \left(\mathrm{iii}\right)\because \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{BAP}=\angle \mathrm{CAP}\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]\\ \Rightarrow \text{\hspace{0.17em}}\mathrm{AP}\text{bisects}\angle \mathrm{A}.\\ \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}}\mathrm{In}\text{}\mathrm{\Delta BDP}\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{and\Delta}\text{\hspace{0.17em}}\mathrm{CDP}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{BD}=\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}}\left[\mathrm{Given}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{DP}=\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}}\left[\mathrm{Common}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{BP}=\mathrm{CP}\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]\\ \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}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{\Delta BDP}\text{\hspace{0.17em}}\cong \mathrm{\Delta}\text{\hspace{0.17em}}\mathrm{CDP}\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{BDP}=\angle \mathrm{CDP}\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]\\ \Rightarrow \mathrm{DP}\text{bisects\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{BDC}\text{i.e.,}\angle \mathrm{D}.\\ \mathrm{Thus},\text{AP bisects}\angle \mathrm{A}\text{as well as}\angle \mathrm{D}.\\ \left(\mathrm{iv}\right)\text{}\angle \text{APB}=\angle \text{APC\hspace{0.17em}\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]\\ \mathrm{and}\text{\hspace{0.17em}\hspace{0.17em}}\angle \text{APB}+\angle \text{APC}=\text{180\xb0}\\ \Rightarrow \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \text{APB}+\angle \text{APB}=\text{180\xb0}\\ \Rightarrow \text{\hspace{0.17em}\hspace{0.17em}}2\angle \text{APB}=180\mathrm{\xb0}\\ \Rightarrow \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \text{APB}=\frac{180\mathrm{\xb0}}{2}=90\mathrm{\xb0}\\ \text{And BP}=\mathrm{C}\text{P}\\ \text{So},\text{ AP is perpendicular bisector of BC.}\end{array}$

**Q.2 **AD is an altitude of an isosceles triangle ABC in which AB = AC. Show that

(i) AD bisects BC

(ii) AD bisects ∠A.

**Ans**

$\begin{array}{l}\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}Given}:\mathrm{AD}\text{is altitude in isosceles}\mathrm{\Delta}\text{ABC in which AB}=\text{AC.}\\ \mathrm{To}\text{\hspace{0.17em}prove}:\left(\text{i}\right)\text{AD bisects BC}\left(\text{ii}\right)\text{AD bisects\hspace{0.17em}\hspace{0.17em}}\angle \text{A}\\ \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)\mathrm{In}\text{}\mathrm{\Delta}\text{ABD and}\mathrm{\Delta}\text{ACD}\\ \text{AB}=\text{AC\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\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}}\angle \mathrm{ADB}=\angle \mathrm{ADC}\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}[Each 90\xb0]}\\ \text{AD}=\text{AD\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\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}}\mathrm{\Delta}\text{ABD}\cong \mathrm{\Delta}\text{ACD\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{R.H.S.}\right]\\ \text{So},\text{ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}BD}=\text{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}}\left[\mathrm{By}\text{C.P.C.T.}\right]\\ \text{Therefore},\text{AD bisects BC.}\\ \left(\mathrm{ii}\right)\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{BAD}=\angle \mathrm{CAD}\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{C.P.C.T.}\right]\\ \text{Therefore},\text{AD bisects}\angle \mathrm{A}.\text{\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 **

$\begin{array}{l}\mathrm{Two}\mathrm{sides}\mathrm{AB}\mathrm{and}\mathrm{BC}\mathrm{and}\mathrm{median}\mathrm{AM}\mathrm{of}\mathrm{one}\mathrm{triangle}\mathrm{ABC}\mathrm{are}\\ \mathrm{respectively}\mathrm{equal}\mathrm{to}\mathrm{sides}\mathrm{PQ}\mathrm{and}\mathrm{QR}\mathrm{and}\mathrm{median}\mathrm{PN}\mathrm{of}\\ \mathrm{\Delta PQR}(\mathrm{see}\mathrm{figure}\mathrm{below}).\mathrm{Show}\mathrm{that}:\\ \left(\mathrm{i}\right)\mathrm{\Delta ABM}\cong \mathrm{\Delta PQN}\\ \left(\mathrm{ii}\right)\mathrm{\Delta ABC}\cong \mathrm{\Delta PQR}\end{array}$

**Ans**

$\begin{array}{l}\text{Given:In}\Delta \text{ABC and}\Delta \text{PQR,}\text{\hspace{0.17em}}\text{AB}=\text{PQ,}\text{\hspace{0.17em}}\text{BC}=\text{QR and AM}=PN.\\ \mathrm{To}\text{prove:}\text{\hspace{0.17em}}\left(i\right)\Delta ABM\cong \Delta PQN\\ \left(ii\right)\Delta ABC\cong \Delta PQR\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{Proof}:\text{(i)In}\mathrm{\Delta}\text{ABM and}\Delta \text{PQN}\text{\hspace{0.17em}}\text{}\\ \text{AB}=\text{PQ}\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[\mathrm{Given}\right]\text{\hspace{0.17em}}\\ \text{BM}=\text{QN}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}[\frac{1}{2}BC=\frac{1}{2}QR]\\ \text{AM}=PN\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[Given\right]\\ \therefore \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\Delta \text{ABC}\cong \Delta \text{PQR}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left[By\text{S}\text{.S}\text{.S}\text{.}\right]\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\angle ABM=\angle PQN\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left[By\text{C}\text{.P}\text{.C}\text{.T}\text{.}\right]\\ \Rightarrow \angle B=\angle Q\\ \left(ii\right)\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}In\text{}\Delta \text{ABC and}\Delta \text{PQR}\\ \text{AB}=\text{PQ}\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[\text{Given}\right]\text{\hspace{0.17em}}\\ \angle B=\angle Q\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[\mathrm{Proved}\text{above}\right]\\ \text{BC}=\text{QR}\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[\text{Given}\right]\text{\hspace{0.17em}}\\ \therefore \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\Delta ABC\cong \Delta PQR\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left[By\text{S}\text{.A}\text{.S}\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}}\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}\hspace{0.17em}Hence Proved}\text{.}\end{array}$

**Q.4** BE and CF are two equal altitudes of a triangle ABC.

Show that:

(i) ΔABE ≅ ΔACF

(ii) AB = AC, i.e ABC is an isosceles triangle.

**Ans**

$\begin{array}{l}\text{Given}:\mathrm{In}\text{}\mathrm{\Delta}\text{ABC, BE}\perp \text{AC and CF}\perp \text{AB and BE = CF}\\ \text{To prove:}\mathrm{\Delta}\text{ABC is isosceles triangle.}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Proof:(i)\hspace{0.33em}}\mathrm{In}\text{}\mathrm{\Delta}\text{BCE and}\mathrm{\Delta}\text{CBF}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{BEC}=\angle \mathrm{CFB}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{Each}\text{90\xb0}\right]\\ \text{\hspace{0.17em}}\mathrm{BC}=\mathrm{BC}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\text{Common}\right]\\ \mathrm{BE}=\mathrm{CF}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}[\text{Given]}\\ \therefore \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{\Delta}\text{BCE}\cong \mathrm{\Delta}\text{CBF\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{By}\text{R.H.S.}\right]\\ \left(\mathrm{ii}\right)\mathrm{AB}=\mathrm{AC}[\mathrm{By}\mathrm{CPCT}]\\ \text{i.e ABC is an isosceles triangle}\end{array}$

**Q.5 **ABC is an isosceles triangle with AB = AC. Draw AP ⊥ BC to show that ∠B = ∠C.

**Ans**

$\begin{array}{l}\text{Given:In}\mathrm{\Delta}\text{ABC, AB}=\text{AC and AP}\perp \text{BC}\\ \text{To prove:}\mathrm{\Delta}\text{ABC is isosceles triangle.}\\ \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}\text{ABP and}\mathrm{\Delta A}\text{CP}\\ \angle \mathrm{APB}=\angle \mathrm{APC}\text{\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\xb0}\right]\\ \text{AB}=\text{AC\hspace{0.17em}\hspace{0.17em}\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{AP}=\text{AP\hspace{0.17em}\hspace{0.17em}\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}}\mathrm{\Delta}\text{ABP}\cong \mathrm{\Delta A}\text{CP\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{R.H.S.}\right]\\ \therefore \angle \mathrm{ABP}=\angle \mathrm{ACP}\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]\\ \Rightarrow \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{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}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\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}$

## FAQs (Frequently Asked Questions)

### 1. What are the characteristics of a triangle?

The three-sided polygonal shape is known as a triangle. It is a two-dimensional, closed shape made of only straight lines. As a result, every Triangle has three vertices in total, and each one makes an angle. The following are the primary properties of a triangle.

- It has three sides, which results in three angles, the sum of which is 180 degrees.
- A Triangle has 360 degrees when its exterior angles are added together.
- A Triangle’s third side is always longer than the length of any combined two of its other sides.
- The Triangle’s shortest side is always located across from its smallest interior angle.

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One of the most important areas of mathematics is geometry, which includes trigonometry, which deals with the sides and angles of a right triangle. It has a wide range of uses in the building and architectural industries.

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