In an isosceles ΔABC with AC = BC, the base AB is produced both ways to P and Q such that AP x BQ = AC². Prove that : ΔACP ~ ΔBQC

Question

In an isosceles ΔABC with AC = BC, the base AB is produced both ways to P and Q such that AP x BQ = AC2. Prove that : ΔACP ~ ΔBQC

Philoid · Accepted Answer

Given ABC is an isosceles triangle and AC = BC

∵ AC = BC

⇒∠CAB = ∠CBA

⇒180° – ∠CAB = 180° – ∠CBA

⇒ ∠CAP = ∠CBQ

Also, AP x BQ = AC²

(∵ AC =BC)

Thus, by SAS similarity, we get

ACP ~ BQC

Hence Proved