Question

prove that"angle opposite to equal side of an isisceles 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.

Answer 2

Answer:

$it \: is \: your \: proof \: \: \: \: .............$