Question

Q.10 Prove that Angles opposite to equal sides of an
isosceles triangle are equal.
​

Answer 1

Step-by-step explanation:

Diagram:-

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Given:-

$\sf \triangle ABC\:is\:a\;isosceles\:triangle$

Where AB=AC

To prove:-

$\sf \angle B=\angle C$

Construction:-

Draw the bisector of $\angle A$ which intersect BC at D

Proof:-

$\sf in\;\triangle ABD \:and\triangle ACD,$

$\because \begin {cases}\sf AB=AC \quad (Given) \\ \sf \angle BAD=\angle CAD\quad (By\:construction) \\ \sf AD=AD \quad (Common)\end {cases}$

$\\\\\therefore \sf \triangle ABD\cong \triangle ACD\quad {}_{\left [Side-Angle-Side(SAS)\right]}$

Hence

$\sf\angle B=\angle C\qquad_{(proved)}$