Answer:
Explanation:
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]