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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 March 2017, 7:51 PM
Answered by Expert
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
Answered by Expert 24th March 2017, 10:27 PM
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