# 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 Fig. 9.26). Show that ar (ABCD) = ar (PBQR).

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

**Solution:**

Let us join AC and PQ.

ΔACQ and ΔAQP are lying on the same base AQ and existing between the same parallels AQ and CP.

According to Theorem 9.2: Two triangles on the same base (or equal bases) and between the same parallels are equal in area.

Area (ΔACQ) = Area (ΔAPQ)

Subtracting ar (ΔABQ) on both sides.

ar (ΔACQ) - ar (ΔABQ) = ar (ΔAPQ) - ar (ΔABQ)

ar (ΔABC) = ar (ΔQBP) ... (1)

Since AC and PQ are diagonals of parallelograms ABCD and PBQR respectively,

Therefore, ar (ΔABC) = 1/2 ar (ABCD)... (2)

Similarly, ar (ΔQBP) = 1/2 ar (PBQR )... (3)

From Equations (1), (2), and (3), we obtain

1/2 ar (ABCD) = 1/2 ar (PBQR)

ar (ABCD) = ar (PBQR) proved.

**Video Solution:**

## 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 Fig. 9.26). Show that ar (ABCD) = ar (PBQR). [Hint: Join AC and PQ. Now compare ar (ACQ) and ar (APQ).]

### Maths NCERT Solutions Class 9 - Chapter 9 Exercise 9.3 Question 9:

**Summary:**

If 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, then ar (ABCD) = ar (PBQR).