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

Given : In ΔABC , CA – CB and AP x BQ = AC²

To prove :- ΔAPC ~ BCQ

Proof:-

AP X BQ = AC² (Given)

Or, AP x BC = AC x AC

Or, AP x BC = AC x BC (AC = BC given)

Or, AP/BC = AC/PQ ………………(i)

Since, CA = CB (Given)

Then, <CAB = <CBA …………….(ii) (Opposite angle to equal sides)

NOW, <CAB +<CAP = 180° …………(iii) (Linear pair of angle)

And <CBA + <CBQ = 180° …………..(iv) (Linear pair of angle)

Compare equation (ii) (iii) & (iv)

<CAP = <CBQ……………..(v)

In ΔAPC and ΔBCQ

<CAP = < CBQ (From equation v)

AP/BC = AC/PQ (From equation i)

Then , ΔAPC ~ ΔBCQ (By SAS similarity)