Question

Pls answer 2nd part

Answer 1

Step-by-step explanation:

In ∆ABC and ∆ACP

AB=AC (since ∆ABC is isosceles)

AP=AP (common side)

BAP=PAC (from I)

∆ABP≈ ∆ACP (SAS)

therefore BP=PC (c.p.c.t)

APB= APC

APB+APC =180° (angles in linear pair)

therefore APB=APC (from above)

APB=APC =180°/ 2=90°

BP=PC and APB=APC=90°

Hence AP is perpendicular Bisector of BC.