ABC is an isosceles triangle in which AB = AC. P is any point in the interior of △ABC such that ∠ABP = ∠ACP. Prove that

(a) BP = CP

(b) AP bisects ∠BAC.

Below figure shows the isosceles triangle ABC with the points marked:

(a) Given,

AB = AC

∴ ∠B = ∠C ……..(i)

Given, ∠ABP = ∠ACP ……..(ii)

Subtracting (ii) from (i) we get,

∠B - ∠ABP = ∠C - ∠ACP

∠PBC = ∠PCB.

∴ BP = CP (As sides opposite to equal angles are equal)

Hence, proved that BP = CP.

(b) We know that,

BP = CP (Proved)

AB = AC (Given)

∠ABP = ∠ACP (Given)

Hence, △ABP ≅ △ACP by SAS axiom.

∠PAB = ∠PAC (Corresponding angles of congruent triangles are equal.)

Thus AP, bisects ∠BAC.

Hence, proved that AP bisects ∠BAC.