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
Plzzz answer this question.....
Its urgent....
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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
Plzzz answer this question.....
Its urgent....
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Answer:
In ΔABC, ∠ABC = 2∠ACB
Let ∠ACB = x
⇒∠ABC = 2∠ACB = 2x
Given BP is bisector of ∠ABC
Hence ∠ABP = ∠PBC = x
By angle bisector theorem the bisector of an angle divides the side opposite to it in the ratio of other two sides.
Hence AB:BC = CP:PA
2) Consider ΔABC and ΔAPB
∠ABC = ∠APB [Exterior angle property]
∠BCP = ∠ABP [Given]
∴ ΔABC ≈ ΔAPB [AA criterion]
∴ AB / BP = CA / CB [Corresponding sides of similar triangles are proportional.]
⇒ AB x BC = BP x CA.