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:
(i) CB : BA = CP : PA
(ii) AB x BC = BP X CA​

Answer 1

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1) In ΔABC, ∠ABC = 2∠ACB

Let ∠ACB = x

⇒∠ABC = 2∠ACB = 2x

Given BP is bisector of ∠ABC

Hence ∠ABP = ∠PBC = x

Using the angle bisector theorem, that is,

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

Hence, CB : BA= CP:PA.

2) Consider ΔABC and ΔAPB

∠ABC = ∠APB [Exterior angle property]

∠BCP = ∠ABP [Given]

∴ ΔABC ≈ ΔAPB [AA criterion]

∴fraction numerator space AB over denominator BP end fraction space equals space CA over CB[Corresponding sides of similar triangles are proportional.]

$⇒ AB x BC = BP x CA$