# 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.

### Asked by Sushanta | 24th Mar, 2017, 07:51: PM

Expert Answer:

###
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]

∴[Corresponding sides of similar triangles are proportional.]

⇒ AB x BC = BP x CA

Let ∠ACB = x

⇒∠ABC = 2∠ACB = 2x

Given BP is bisector of ∠ABC

Hence ∠ABP = ∠PBC = x

Using the angle bisector theorem, that is,

Hence, CB : BA= CP:PA.

2) Consider ΔABC and ΔAPB

∠ABC = ∠APB [Exterior angle property]

∠BCP = ∠ABP [Given]

∴ ΔABC ≈ ΔAPB [AA criterion]

∴[Corresponding sides of similar triangles are proportional.]

⇒ AB x BC = BP x CA

### Answered by Rebecca Fernandes | 24th Mar, 2017, 10:27: PM

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