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 

Answer 1

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