Answer :

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)


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