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 CB:BA=CP:PA

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