# NCERT Solutions for Class 9 Maths Chapter 7 Triangles (Ex 7.2) Exercise 7.2

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## NCERT Solutions for Class 9 Maths Chapter 7 Triangles (Ex 7.2) Exercise 7.2

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### Access NCERT Solutions for Class 9 Maths Chapter 7 – Triangles

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### NCERT Solutions for Class 9 Maths Chapter 7 Triangles Exercise 7.2

Significant characteristics of Triangles and associated theorems are covered in NCERT Solutions For Class 9 Maths Chapter 7 Exercise 7.2. Due to the fact that this topic involves more proving than solving, a majority of students find it tedious. Students will begin to show interest in it if they can relate the characteristics of Triangles to the examples and comprehend them. For the convenience of students, Extramarks has provided the NCERT Solutions For Class 9 Maths Chapter 7 Exercise 7.2, on their website and learning app. Students may quickly comprehend the process for answering the problems after reading the solutions. Extramarks’ skilled teachers have prepared the NCERT Solutions For Class 9 Maths Chapter 7 Exercise 7.2.

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### NCERT Solutions for Class 9

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All the other chapters’ NCERT solutions for Class 9 are also available on the Extramarks website and mobile application.

Chapter 1 Number System

Chapter 2 Polynomials

Chapter 3 Coordinate Geometry

Chapter 4 Linear Equations in Two Variables

Chapter 5 Introduction to Euclids Geometry

Chapter 6 Lines and Angles

Chapter 7 Triangles

Chapter 9 Areas of Parallelograms and Triangles

Chapter 10 Circles

Chapter 11 Constructions

Chapter 12 Heron’s Formula

Chapter 13 Surface Areas and Volumes

Chapter 14 Statistics

Chapter 15 Probability

### CBSE Study Materials for Class 9

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### CBSE Study Materials

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Q.1 In an isosceles triangle ABC, with AB = AC, the bisectors of ∠B and ∠C intersect each other at O. Join A to O. Show that :
(i) OB = OC
(ii) AO bisects ∠A.

Ans

$\begin{array}{l}\text{Given: In}\mathrm{\Delta }\text{ABC, AB}=\text{AC and bisectors of}\angle \text{B and}\angle \text{C intersect}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}at O.}\\ \text{To Prove:}\left(\text{i}\right)\text{OB}=\text{OC}\left(\text{ii}\right)\text{AO bisects}\angle \text{A}\\ \text{Proof:}\left(\mathrm{i}\right)\text{In}\mathrm{\Delta }\text{ABC, AB}=\text{AC}\\ \text{So, \hspace{0.17em}\hspace{0.17em}}\angle \mathrm{ACB}=\angle \mathrm{ABC}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\begin{array}{l}\text{Opposite angles of equal}\\ \text{sides are equal.}\end{array}\right]\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{1}{2}\angle \mathrm{ACB}=\frac{1}{2}\angle \mathrm{ABC}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{OCB}=\angle \mathrm{OBC}\text{\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\begin{array}{l}\text{Given},\text{as OB and OC are angle}\\ \text{bisectors.}\end{array}\right]\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{OB}=\mathrm{OC}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em} \hspace{0.17em}}\left[\begin{array}{l}\text{Opposite angles of equal}\\ \text{sides are equal.}\end{array}\right]\end{array}$

$\begin{array}{l}\left(\mathrm{ii}\right)\text{In}\mathrm{\Delta O}\text{AB and}\mathrm{\Delta O}\text{AC,}\\ \text{\hspace{0.17em}\hspace{0.17em}}\mathrm{OA}=\mathrm{OA}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{Common}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}}\mathrm{AB}=\mathrm{AC}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{Given}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}}\mathrm{OB}=\mathrm{OC}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\text{Proved above}\right]\\ \therefore \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{\Delta O}\text{AB}\cong \text{}\mathrm{\Delta O}\text{AC\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{By}\text{S.S.S.}\right]\\ \mathrm{Hence},\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{OAB}=\angle \mathrm{OAC}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{By}\text{C.P.C.T.}\right]\\ \mathrm{Thus},\text{OA bisects}\angle \text{A.}\end{array}$

Q.2 In Δ ABC, AD is the perpendicular bisector of BC (see figure below). Show that Δ ABC is an isosceles triangle in which AB = AC.

Ans

$\begin{array}{l}\text{Given}:\text{In}\mathrm{\Delta }\text{ABC, BD}=\text{DC and AD}\perp \text{BC.}\\ \text{To Prove: AB}=\text{AC}\\ \text{Proof: In}\mathrm{\Delta }\text{ABD and}\mathrm{\Delta }\text{ACD}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{AD}=\mathrm{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}\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}}\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}\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}}\mathrm{B}\mathrm{D}=\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}}\left[\text{Given]}\\ \therefore \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{\Delta }\text{ABD}\cong \text{}\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}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{By}\text{S.A.S.}\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{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}\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{}\mathrm{\Delta }\text{ABC is isosceles triangle.}\end{array}$

Q.3 ABC is an isosceles triangle in which altitudes BE and CF are drawn to equal sides AC and AB respectively (see figure below). Show that these altitudes are equal.

Ans

$\begin{array}{l}\text{Given:In\hspace{0.17em}}\mathrm{\Delta ABC},\text{\hspace{0.17em}}\mathrm{AC}=\mathrm{AB}\text{, BE}\perp \text{AC and CF}\perp \text{AB.}\\ \text{To prove: BE}=\text{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}Proof:}\mathrm{In}\text{\hspace{0.17em}}\mathrm{\Delta ABE}\text{and\hspace{0.17em}}\mathrm{\Delta ACF}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{BAE}=\angle \mathrm{CAF}\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}[Common angle]}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{BEA}=\angle \mathrm{CFA}\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]\\ \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}\hspace{0.17em}\hspace{0.17em}}\left[\text{Given]}\\ \therefore \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{\Delta ABE}\cong \text{\hspace{0.17em}}\mathrm{\Delta ACF}\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{By}\text{A.A.S.}\right]\\ \therefore \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}}\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}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\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

$\begin{array}{l}\mathrm{ABC}\mathrm{is}\mathrm{a}\mathrm{triangle}\mathrm{in}\mathrm{which}\mathrm{altitudes}\mathrm{BE}\mathrm{and}\mathrm{CF}\mathrm{to}\mathrm{sides}\mathrm{AC}\\ \mathrm{and}\mathrm{AB}\mathrm{are}\mathrm{equal}\left(\mathrm{see}\mathrm{figure}\mathrm{below}\right).\mathrm{Show}\mathrm{that}\\ \left(\mathrm{i}\right)\text{ }\mathrm{\Delta ABE}\cong \mathrm{\Delta ACF}\\ \left(\mathrm{ii}\right)\text{ }\mathrm{AB}=\mathrm{AC},\mathrm{i}.\mathrm{e}.,\mathrm{ABC}\mathrm{is}\mathrm{an}\mathrm{isosceles}\mathrm{triangle}.\end{array}$

Ans

$\begin{array}{l}\mathrm{ }\mathrm{Given}:\mathrm{In}\mathrm{ }\mathrm{\Delta ABC},\mathrm{ }\mathrm{BE}=\mathrm{CF},\mathrm{BE}\perp \mathrm{AC}\mathrm{and}\mathrm{CF}\perp \mathrm{AB}\mathrm{.}\\ \mathrm{To}\mathrm{prove}:\left(\mathrm{i}\right)\text{ }\mathrm{\Delta ABE}\cong \mathrm{\Delta ACF}\\ \left(\mathrm{ii}\right)\text{ }\mathrm{AB}=\mathrm{AC},\mathrm{i}.\mathrm{e}.,\mathrm{ABC}\mathrm{is}\mathrm{an}\mathrm{isosceles}\mathrm{triangle}.\\ \mathrm{Proof}:\left(\mathrm{i}\right)\text{ }\mathrm{In}\mathrm{ }\mathrm{\Delta ABE}\mathrm{and}\mathrm{ }\mathrm{\Delta ACF}\\ \angle \mathrm{BAE}=\angle \mathrm{CAF} \mathrm{ }\left[\mathrm{Common}\mathrm{angle}\right]\\ \angle \mathrm{BEA}=\angle \mathrm{CFA} \mathrm{ }\left[\mathrm{Each}90°\right]\\ \mathrm{BE}=\mathrm{CF} \mathrm{ } \mathrm{ }\left[\mathrm{Given}\right]\\ \therefore \mathrm{\Delta ABE}\cong \mathrm{ }\mathrm{\Delta ACF} \left[\mathrm{By}\mathrm{A}.\mathrm{A}.\mathrm{S}\mathrm{.}\right]\\ \left(\mathrm{ii}\right)\mathrm{ }\mathrm{AB}=\mathrm{AC} \left[\mathrm{By}\mathrm{C}.\mathrm{P}.\mathrm{C}.\mathrm{T}\mathrm{.}\right]\\ \mathrm{i}.\mathrm{e}.,\mathrm{ABC}\mathrm{is}\mathrm{an}\mathrm{isosceles}\mathrm{triangle}.\\ \mathrm{Hence}\mathrm{proved}\mathrm{.}\end{array}$

Q.5 ABC and DBC are two isosceles triangles on the same base BC (see figure below). Show that ∠ABD = ∠ACD.

Ans

$\begin{array}{l}\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{Given}:\mathrm{In}\text{\hspace{0.17em}}\mathrm{\Delta ABC},\text{\hspace{0.17em}}\mathrm{AB}=\mathrm{AC}\text{and}\mathrm{in}\text{\hspace{0.17em}}\mathrm{\Delta DBC},\text{\hspace{0.17em}}\mathrm{DB}=\mathrm{DC}\text{.}\\ \text{To prove:}\angle \mathrm{ABD}=\angle \mathrm{ACD}\\ \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:}\mathrm{In}\text{\hspace{0.17em}}\mathrm{\Delta ABC}\\ \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}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{Given}\right]\\ \therefore \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{ACB}=\text{\hspace{0.17em}}\angle \mathrm{ABC}\text{\hspace{0.17em}\hspace{0.17em}}...\left(\mathrm{i}\right)\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\begin{array}{l}\mathrm{Opposite}\text{angles of equal sides}\\ \text{are equal.}\end{array}\right]\\ \mathrm{In}\text{\hspace{0.17em}}\mathrm{\Delta DBC}\\ \mathrm{DB}=\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}}\left[\mathrm{Given}\right]\\ \therefore \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{DCB}=\text{\hspace{0.17em}}\angle \mathrm{DBC}\text{\hspace{0.17em}\hspace{0.17em}}...\left(\mathrm{ii}\right)\text{\hspace{0.17em}\hspace{0.17em}}\left[\begin{array}{l}\mathrm{Opposite}\text{angles of equal sides}\\ \text{are equal.}\end{array}\right]\\ \mathrm{Adding}\text{equation}\left(\mathrm{i}\right)\text{and equation}\left(\mathrm{ii}\right),\mathrm{we}\text{}\mathrm{get}\\ \angle \mathrm{ACB}\text{\hspace{0.17em}}+\angle \mathrm{DCB}=\text{\hspace{0.17em}}\angle \mathrm{ABC}+\angle \mathrm{DBC}\\ \therefore \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{ACD}=\text{\hspace{0.17em}}\angle \mathrm{ABD}\\ ⇒\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}}\angle \mathrm{ABD}=\angle \mathrm{ACD}\\ \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}}\mathrm{Hence}\text{proved.}\end{array}$

Q.6 ΔABC is an isosceles triangle in which AB = AC. Side BA is produced to D such that AD = AB (see figure below). Show that ∠BCD is a right angle.

Ans

$\begin{array}{l}\text{Given:In​}\mathrm{\Delta }\text{ABC, AB}=\text{AC, side BA​ is produced to D such that}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}AD}=\text{AB.}\\ \text{To prove:}\angle \text{BCD is a right angle}.\\ \text{Proof:In}\mathrm{\Delta }\text{ABC,}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}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}}\left[\text{Given]}\\ \angle \text{ACB}=\angle \text{ABC \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\text{Opposite angles of equal sides are equal.}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\mathrm{x}\text{\hspace{0.17em}\hspace{0.17em}}\left(\mathrm{let}\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 }\text{ACD,}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}AC}=\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}[Given]}\\ \angle \text{ADC}=\angle \text{ACD \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\text{Opposite angles of equal sides are equal.}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=\mathrm{y}\text{\hspace{0.17em}\hspace{0.17em}}\left(\mathrm{let}\right)\\ \text{Now},\text{\hspace{0.17em}In}\mathrm{\Delta }\text{BCD}\\ \angle \text{BCD}+\angle \mathrm{C}\text{BD}+\angle \text{BDC}=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}\hspace{0.17em}\hspace{0.17em}}\left(\mathrm{x}+\mathrm{y}\right)+\mathrm{x}+\mathrm{y}=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}}2\left(\mathrm{x}+\mathrm{y}\right)=180\mathrm{°}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left(\mathrm{x}+\mathrm{y}\right)=\frac{180\mathrm{°}}{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \text{BCD}=90\mathrm{°}\\ ⇒∆\text{BCD is right triangle.}\end{array}$

Q.7 ABC is a right angled triangle in which ∠A = 90° and AB = AC. Find ∠B and ∠C.

Ans

$\begin{array}{l}\text{\hspace{0.17em}\hspace{0.17em}Given :In}\mathrm{\Delta }\text{ABC,}\angle \text{A}=\text{90° and AB}=\text{AC.}\\ \text{To find :\hspace{0.17em}}\angle \text{\hspace{0.17em}B and}\angle \text{\hspace{0.17em}C}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Sol : In}\mathrm{\Delta }\text{ABC,}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{AB}=\mathrm{AC}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\text{Given]}\\ \mathrm{So},\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{ACB}=\angle \mathrm{ABC}=\mathrm{x}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left(\mathrm{let}\right)\\ \text{Since},\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{A}+\angle \mathrm{B}+\angle \mathrm{C}=180\mathrm{°}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}90\mathrm{°}+\mathrm{x}+\mathrm{x}=180\mathrm{°}\\ 2\mathrm{x}=180\mathrm{°}-90\mathrm{°}\\ \text{\hspace{0.17em}\hspace{0.17em}}\mathrm{x}=\frac{90\mathrm{°}}{2}=45\mathrm{°}\\ \mathrm{So},\text{\hspace{0.17em}}\angle \mathrm{B}=45\mathrm{°}\text{and}\angle \mathrm{C}=45\mathrm{°}.\end{array}$

Q.8 Show that the angles of an equilateral triangle are 60° each.

Ans

$\begin{array}{l}\mathrm{Given}:\mathrm{\Delta ABC}\text{is an equilateral triangle.}\\ \text{To prove :}\angle \text{A}=\angle \mathrm{B}=\angle \text{C}=60\mathrm{°}\\ \text{Proof}:\text{In}\mathrm{\Delta ABC},\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{AB}=\mathrm{AC}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{Given}\right]\end{array}$

$\begin{array}{l}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{C}=\angle \mathrm{B}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\left(\mathrm{i}\right)\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\begin{array}{l}\text{Angles opposite to equal sides}\\ \text{are equal.}\end{array}\right]\\ \text{\hspace{0.17em}}\mathrm{AB}=\mathrm{BC}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{Given}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{C}=\angle \mathrm{A}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\left(\mathrm{ii}\right)\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\begin{array}{l}\mathrm{Angles}\text{opposite to equal sides}\\ \text{are equal.}\end{array}\right]\\ \text{By angle sum property in a triangle,}\\ \angle \mathrm{A}+\angle \mathrm{B}+\angle \mathrm{C}=180\mathrm{°}\\ \angle \mathrm{C}+\angle \mathrm{C}+\angle \mathrm{C}=180\mathrm{°}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\text{From equation}\left(\mathrm{i}\right)\text{and equation}\left(\mathrm{ii}\right)\right]\\ 3\angle \mathrm{C}=180\mathrm{°}\\ \text{\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{C}=\frac{180\mathrm{°}}{3}=60\mathrm{°}\\ \mathrm{So},\text{}\angle \mathrm{A}\text{\hspace{0.17em}}=60\mathrm{°}\text{and}\angle \mathrm{B}\text{\hspace{0.17em}}=60\mathrm{°}\\ \text{Therefore},\text{each angle of an equilateral triangle is 60° each.}\end{array}$