Question

P is a point on the side BC of ΔABC such that ∠APC = ∠BAC. Prove that AC² = BC · CP.

Answer 1

In ΔBAC and ΔΑBC, ΔΑΡΟ

∠BAC = ∠APC  .....(Given)

∠ACB = ∠ACP  .....(Common)

∴ ΔВАС ∼ ΔАРС   ....(AA similarity)

∴ `(BA)/(AP) = (AC)/(PC) = (BC)/(AC)` 

Take `(AC)/(PC) = (BC)/(AC)`

AC² = BC · CP

Hence proved.