**The side AB of a parallelogram ABCD is produced to any point P. A line through A and parallel to CP meets CB produced at Q and then parallelogram PBQR is completed (see figure). Show that ar(ABCD)=ar(PBQR).**

[**Hint:** Join AC and PQ. Now compare ar(ACQ) and ar(APQ)]

**Solution:**

It is given that ABCD and PBQR are two parallelograms.

We need to show that ar(ABCD)=ar(PBQR)

**Construction: **Join AC and PQ.

We have

**(Triangles on the same base and between the same parallels are equal in area.)**

**(Diagonal of parallelogram divides it into two triangles of equal areas.)**