Question

In an isosceles ∆ABC, the base AB is produced both ways in P and Q such that 

AP × BQ = AC² . 

Prove that ∆ACP~∆BCQ.

Answer 1

Disclaimer: It should be ∆APC ~ ∆BCQ instead of ∆ACP ~ ∆BCQ 

It is given that ∆ABC is an isosceles triangle. 

Therefore, 

CA = CB 

⇒ ∠CAB = ∠CBA

⇒ 1800 − ∠CAB  = 1800 − ∠CBA

⇒ ∠CAP = ∠CBQ 

Also, 

AP × BQ = AC² 

⇒ AP/AC= AC/BQ

⇒ AP/AC = BC/BQ (∵ AC = BC) 

Thus, by SAS similarity theorem, we get 

∆APC ~ ∆BCQ 

This completes the proof