Answer :

In a ΔAXP and ΔCXB,

PAX = XCB (Alternative angles AP || BC)


AX = CX (Given)


AXP = CXB (Vertically opposite angles)


ΔAXP ΔCXB (By ASA rule)


AP = BC (By c.p.c.t) (i)


Similarly,


QA = BC (ii)


From (i) and (ii), we get


AP = QA


Now,


AP || BC


And,


AP = QA


Area (APB) = Area (ACQ) (Therefore, Triangles having equal bases and between the same parallels QP and BC)


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