Question

In triangle ABC, angle ABC is equal to twice the angle ACB, and bisector of angle ABC meets the opposite side at point P. Show that: (1) CB : BA = CP : PA (2) AB * BC = BP * CA

Answer 1

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.

Answer 2

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