Question

5. 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​

Answer 1

Step-by-step explanation:

In △ABC

∠ABC=2∠ACB ( given)

Let ∠ACB=x

BP is bisector of ∠ABC

∴∠ABP=∠PBC=x

By angle bisector theorem

the bisector of an divides the side opposite to it in ratio of other two sides

Hence,

$\frac{ab}{bc}$

$= \frac{cb}{\ ba} = \frac{cp}{pa}$

Consider △ABC and △APB

∠ABC=∠APB (exterior angle properly)

given that, ∠BCP=∠ABP

∴ABC≈△APB (AA criteria)

$\frac{ab}{bp } = \frac{ca}{cb}$

∴AB×BC=BP×CA

HENCE PROVED