In the given figure, ABCD and ABPQ are two parallelograms and M is a point on AQ and BMP is a triangle.

Then, ar(∆BMP) = ar(gm ABCD).


We know that any two or parallelogram having the same base and lying between the same parallel lines are equal in area.


ar(||gm ABCD) = ar(||gm ABPQ) –1


We also know that when a parallelogram and a triangle lie on same base and between same parallel lines then, area of the triangle is half the area of the parallelogram.


ar(∆BMP) = ar(gm ABPQ)


But, from –1


ar(||gm ABCD) = ar(||gm ABPQ)


ar(∆BMP) = ar(gm ABCD)

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