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ICSE Class 10 Answered

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 | 24 Mar, 2017, 07:51: PM
answered-by-expert 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]

∴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 Rebecca Fernandes | 24 Mar, 2017, 10:27: PM

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