In ΔABC, 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

ΔABC is shown in the figure below:

(i) Let ∠ACB = x, so ∠ABC = 2x.

Since, BP is the bisector of ∠ABC.

So, ∠ABP = ∠PBC = x.

By angle bisector theorem,

The internal bisector of an angle of a triangle divides the opposite side internally in the ratio of the sides containing the angle.

$\therefore \dfrac{CB}{BA} = \dfrac{CP}{PA}$

i.e. CB : BA = CP : PA.

Hence, proved that CB : BA = CP : PA.

(ii) From figure,

∠APB = ∠PBC + ∠PCB = 2x. [Exterior angle is equal to the sum of opposite two interior angles].

∴ ∠APB = ∠ABC

∠BCP = ∠ABP [Both = x]

∴ △ABC ~ △APB [By A.A.]

Since corresponding sides of similar triangles are proportional we have,

⇒ $\dfrac{CA}{AB} = \dfrac{BC}{BP}$

⇒ AB x BC = BP x CA

Hence, proved that AB x BC = BP x CA.