Answer :


Given, CD is parallel to PQ, Two Circles with centers P and Q intersect at point A and B.


To Prove: 2PQ = CD


Construction: PE and QE are the Perpendiculars on the Chord CD From centers P and Q respectively.


DE = EA


CF = AF


DE + EA + AF + FC = DC


2EA + 2AF = DC


2(EA + AF) = DC ……..(1)


PQ and CD are parallel lines and PE and QE are the perpendiculars.


So, AEP = 900,AFQ = 900,EPQ = 900,AFQ = 900


EPQF is a rectangle


So, EF = PQ ……… (2)


2(EA + AF) = DC


2EF = DC


2PQ = DC


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