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)