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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.
Solution
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.
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