Question

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

Answer 1

We are given a triangle ΔABC with a point P on BC such that:

∠APC = ∠BAC

AC² = BC·CP

Since ∠APC = ∠BAC, we use the Angle-Angle (AA) similarity criterion: 

△APC ∼ △BAC

(Since they have one common angle ∠ACB and ∠APC = ∠BAC).

For similar triangles, the ratio of corresponding sides is equal:

`(AC)/(BC) = (PC)/(AC)` 

Cross multiply: AC² = BC·CP

Hence, proved.