In the give figure, X and Y are the midpoints of AC and AB respectively, QP ‖ BC and CYQ and BXP are straight lines. Prove that ar(∆ABP) = ar(∆ACQ).

Given: X and Y are the midpoints of AC and AB respectively, QP BC and CYQ and BXP are straight lines.

Construction: Join QB and PC

Area (Δ QBC) = Area (Δ BCP) (Triangles on same base BC and between same parallel lines are equal in area) –1 and,

Area (||gm ACBQ) = Area (||gm ABCP) (parallelograms on same base BC and between same parallel lines are equal in area) –2

Subtract –1 from –2

Area (||gm ACBQ) – Area (Δ QBC) = Area (||gm ABCP) – Area (Δ BCP)

Area (∆ACQ) = Area (∆ABP)

Area(∆ABP) = Area(∆ACQ)

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