Question

In A ABC, angle ABC is equal to twice the
angle ACB, and bisector of angle ABC meets
the opposite side at point P. Show that:
(i) CB : BA = CP : PA
(ii) AB X BC = BP X CA​



Plzzz answer this question.....
Its urgent....​

Answer 1

Answer:

In ΔABC, ∠ABC = 2∠ACB

Let ∠ACB = x

⇒∠ABC = 2∠ACB = 2x

Given BP is bisector of ∠ABC

Hence ∠ABP = ∠PBC = x

By angle bisector theorem the bisector of an angle divides the side opposite to it in the ratio of other two sides.

Hence AB:BC = CP:PA

2) Consider ΔABC and ΔAPB

∠ABC = ∠APB [Exterior angle property]

∠BCP = ∠ABP [Given]

∴ ΔABC ≈ ΔAPB [AA criterion]

∴ AB / BP = CA / CB [Corresponding sides of similar triangles are proportional.]

⇒ AB x BC = BP x CA.