Question

show that in an isosceles triangle the angles opposite to equal sides are equal
​

Answer 1

Take a triangle ABC, in which AB=AC.

Construct AP bisector of angle A meeting BC at P.

In ∆ABP and ∆ACP

AP=AP[common]

AB=AC[given]

angle BAP=angle CAP[by construction]

Therefore, ∆ABP congurent ∆ACP[S.A.S]

This implies, angle ABP=angleACP[C.P.C.T]

Hence proved that angles opposite to equal sides of a triangle are equal.

*Please rate it brainliest*

Answer 2

Step-by-step explanation:

Lets take the triangle as ABC and AD as bisector

angle C = angle B (given)

BA = CA (given)

DAB = DAC (as AD is bisector)

by SAS they are congurent

corresponding part of congurent triangle are equal