In an isosceles ΔABC, the base AB is produced both ways in P and Q such that AP × BQ = AC². Prove that ΔACP ~ ΔBCQ.

To prove: ∆ACP ∼ ∆BCQ

Proof:

Given that, ∆ABC is an isosceles triangle. ⇒ AC = BC

Also, if ∆ABC is an isosceles triangle,

then ∠CAB = ∠CBA …(i)

Subtracting it by 180° from both sides, we get

180° - ∠CAB = 180° - ∠CBA

⇒ ∠CAP = ∠CBQ …(ii)

Also, given that AP × BQ = AC × AC

Or

Or [∵ AC = BC] …(iii)

Recollecting equations (i), (ii) and (iii),

By SAS-similarity criteria, we get

∆ACP ∼ ∆BCQ

Hence, proved.