In an isosceles ∆ ABC, the base AB is produced both ways to P and Q, such that AP × BQ = AC².
Prove that ∆ ACP ∼ ∆ BCQ.

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Solution

Disclaimer: It should be $△ A P C ~ △ B C Q$ instead of ∆ACP ∼ ∆BCQ
It is given that $△$ ABC is an isosceles triangle.
Therefore,
CA = CB
$\Rightarrow ∠ C A B = ∠ C B A \Rightarrow 180 ° - ∠ C A B = 180 ° - ∠ C B A \Rightarrow ∠ C A P = ∠ C B Q$

Also,
$A P \times B Q = A C^{2} \Rightarrow \frac{A P}{A C} = \frac{A C}{B Q} \Rightarrow \frac{A P}{A C} = \frac{B C}{B Q} (∵ AC = BC)$

Thus, by SAS similarity theorem, we get:
$△ A P C ~ △ B C Q$
This completes the proof.

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In an isosceles ∆ ABC, the base AB is produced both ways to P and Q, such that AP × BQ = AC2. Prove that ∆ ACP ∼ ∆ BCQ.

In an isosceles ∆ ABC, the base AB is produced both ways to P and Q, such that AP × BQ = AC².
Prove that ∆ ACP ∼ ∆ BCQ.