In this given figure, ABCD is a trapezium in which AB parallel to DC, O is the mid point of BC. Through the point O, a line PQ parallel to AD has been drawn which intersects AB at Q and DC produced at P. Prove that ar(ABCD) = ar(AQPD)
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Dear Student,
Please find below the solution to the asked query:


Given: In trapezium ABCD, ABDC, CO=BO and PQADTo prove: arABCD=arAQPDProof:In BQO and CPO,BO=CO                    GivenQBO=PCO      Alternate interior angle as ABDCBOQ=COP      Vertically opposite angleSo, by ASA congruence criteria,BQOCPOarBQO=arCPOarBQO+arAQOCD=arCPO+arAQOCD arABCD=arAQPD

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FIRST WE HAVE TO PROVE TRIANGLES CPO AND BOQ CONGRUENT
OC=OB
<CPO=<OQB
<COP=<BOQ
BY AAS BOTH TRIANGLES ARE CONGRUENT 
SO THEIR AREA ARE ALSO EQUAL.
NOW ADD AREA OF ADCOQ TO BOTH 
AREA (ABCD)= AR(AQCD)
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SORRY ITS AR(AQPD) AT THE END
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