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 CB:BA=CP:PA
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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
LikhithaAlena:
it is actually cb:ba=cp:pa
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