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Question
In ΔABC, angle ABC is equal to twice the angle ACB, and bisector of angle ABC meets the opposite side at point P. Show that: AB × BC = BP × CA
Solution
Consider ΔABC and ΔAPB
∠ABC = ∠APB ...[Exterior angle property]
∠BCP = ∠ABP ...[Given]
∴ ΔABC ∼ ΔAPB ...[AA criterion for similarity]
`(CA)/(AB) = (BC)/(BP)` ...(Corresponding sides of similar triangles are proportional)
`=>` AB × BC = BP × CA
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