Q. 13

In an isosceles ΔABC with AC = BC, the base AB is produced both ways to P and Q such that AP x BQ = AC2. Prove that : ΔACP ~ ΔBQC


Answer :

Given ABC is an isosceles triangle and AC = BC

AC = BC


⇒∠CAB = CBA


180° – CAB = 180° – CBA


CAP = CBQ


Also, AP x BQ = AC2



( AC =BC)


Thus, by SAS similarity, we get


ACP ~ BQC


Hence Proved


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