Question

in triangle ABC,AB = AC . P is a point in the interior of the triangle such that angle ABP = angle ACP. prove that AP bisects angle BAC.​

Answer 1

Given: ABC is a triangle in which AB=AC. P is any point in the interior of the triangle such that angle ABP=ACP.

To Prove: AP bisects angle BAC

Proof: 

In  tr. APB and tr. APC,

AB = AC [given]

angle ABP = angle ACP [given]

AP = AP [common]

∴

also, 

angle PAB = angle PAC [corresponding angles of congruent triangles]

thus, AP bisects angle BAC.

hence proved.

Answer 2

Answer:

Step-by-step explanation:

Given-Ab=Ac

To prove- angleAbp=angleAcp

Proof- In triangle ABP and triangle ACP

AB=AC(given)

AP=AP(common)

Angle ABP=Angle ACP(given)

That is triangle ABP congruent to Triangle ACP

So by c.p.c.t

Angle BAP=Angle CAP

That is AP bisects angle BAC