Question

prove that angles opposite to equal sides of a 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. *Please mrk it brainliest*  

Answer 2

Given - ABC is a triangle
To prove - angles opposite to equal sides of a triangle are equal
construction- draw a triangle ABC in which AB=AC and draw a angle bisector of angle A
proof - In∆ BAD and ∆CAD
AB=AC. ( given)
angle BAD =angle CAD (by construction)
AD =AD ( common)
∆BAD =~ ∆CAD ( by ASA rule)
so, angle B = angle C. ( by C.P.C.T)
hence proved