ABC is a triangle in which AB = AC. P is any point in its interior such
that. Prove that AP bisects 

The correct answer is: Hence proved

In  ABC,
AB = AC (given)

 ∠B = ∠C(angles opposite to equal sides are equal)

here, ∠ ABP  =  ∠ ACP (given)
From diagram we get,

∠B = ∠ ABP  +  ∠ PBC —-Eq1

∠C = ∠ ACP  +  ∠ PCB—-Eq2
Subtracting Eq1-Eq2 we get,

∠B - ∠C = ∠ ABP  +  ∠ PBC - ∠ ACP - ∠ PCB

Substitute ∠  ABP  =  ∠ ACP and ∠ B  =  ∠ C,

We get ∠ PBC  =  ∠ PCB

In   PBC, ∠ PBC = ∠PCB

Then , PB  = PC (side opposite to equal angles are equal)

We get PB  =  PC(side )

∠ABP  =  ∠ ACP (given)(Angle)

AB =  AC (given)(side)

By SAS rule, 
By congruence , ∠ BAP  =  ∠ CAP then , ∠ BAC is bisected by AP
Hence proved