In the given figure, the side AB of ∥ gm ABCD is produced to a point P. A line through A drawn parallel to CP meets CB produced in Q and the parallelogram PBQR is completed. Prove that : ar (∥ gm ABCD) = ar (∥ gm BPRQ).

Δ AQC and Δ AQP shares the base AQ lies between the same parallels AQ and CP.

ar (ΔAQC) = ar (ΔAQP) ……(1)

Both triangles share a common part, which is Δ AQB. Subtract this area from both sides of Equation (1):

ar (ΔAQC) - ar (ΔAQB) = ar (ΔAQP) - ar (ΔAQB)

ar (ΔABC) = ar (ΔQPB) …….(2)

We know that a diagonal of a parallelogram divides it into two triangles of equal area.

In parallelogram ABCD, AC is the diagonal,

ar(Δ ABC) = $\dfrac{1}{2}$ ar(∥ gm ABCD)

In parallelogram BPRQ, QP is the diagonal,

ar(Δ QPB) = $\dfrac{1}{2}$ ar(∥ gm BPRQ)

Substituting the values in equation (2), we get :

$\dfrac{1}{2}$ ar(∥ gm ABCD) = $\dfrac{1}{2}$ ar(∥ gm BPRQ)

ar(∥ gm ABCD) = ar(∥ gm BPRQ).

Hence, proved that ar(∥ gm ABCD) = ar(∥ gm BPRQ).