Question

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

 

Answer 1

Disclaimer: It should be ΔAPC ~ ΔBCQ instead of ΔACP ~
ΔBCQ
It is given that ΔABC is an isosceles triangle.
Therefore,    

CA = CB
⟹ ∠𝐶𝐴𝐵 = ∠𝐶𝐵𝐴
⟹ 180°− ∠𝐶𝐴𝐵 = 180° − ∠𝐶𝐵𝐴
⟹ ∠𝐶𝐴𝑃 = ∠𝐶𝐵𝑄
Also, 

`APxxBQ=AC^2`  

⇒` (AP)/(AC)=(AC)/(BQ)` 

⇒ `(AP)/(AC)=(BC)/(BQ)`        (∵𝐴𝐶=𝐵𝐶 ) 

Thus, by SAS similarity theorem, we get
ΔAPC ~ ΔBCQ
This completes the proof.