Question

ABC is an isosceles triangle in which AB = AC. P is any point in the interior of triangle ABC such that angle ABP = angle ACP. Prove that:
(b) AP bisects angle BAC.
(a) BP = CP​

Answer 1

Answer:

In ABC, ab=ac(given)

ap=ap(common)

angle abp=angleacp

so, by s.a.s abp is congruent to apc

so,bp=cp(c.p.c.t)

and ap bisect angle bac

Answer 2

Answer:

BP = CP​

AP bisects ∠ BAC

Step-by-step explanation:

ABC is an isosceles triangle in which AB = AC

=> ∠ABC = ∠ACB

P is any point in the interior of triangle ABC such that angle ∠ABP = ∠ ACP

in Δ PBC

∠PBC = ∠ABC - ∠ABP  = ∠ACB - ∠ ACP = ∠PCB

=> PB = PC

or BP = CP

now in Δ APB & ΔAPC

AP = AP  Common

AB = AC

PB = PC

=> Δ APB ≅ ΔAPC

=> ∠BAP = ∠CAP

∠BAC = ∠BAP + ∠CAP

=> ∠BAP  = ∠CAP = (1/2) ∠BAC

=> AP bisects ∠BAC