Question

prove that angles opposite to equal sides of an isosceles triangle 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.

$\huge\underline\mathbb\blue{Answer}$

$<marquee>$

$\huge\red{\mid{\fbox{\tt{ITZ\: DARE\:DAVIL}}\mid}}$

Answer 2

Answer:

Step-by-step explanation:

hope its usefull