# In Fig. 9.32, ABCD is a parallelogram and BC is produced to a point Q such that AD = CQ. If AQ intersect DC at P, show that ar (BPC) = ar (DPQ). [Hint : Join AC.]

**Solution:**

It is given that ABCD is a parallelogram.

AD || BC and AB || DC (Opposite sides of a parallelogram are parallel to each other).

Now, join the points A and C.

Consider ΔAPC and ΔBPC

ΔAPC and ΔBPC are lying on the same base PC and between the same parallels PC and AB.

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

Therefore,

Area (ΔAPC) = Area (ΔBPC) ...(1)

In quadrilateral ACQD, it is given that AD = CQ

Since ABCD is a parallelogram,

AD || BC (Opposite sides of a parallelogram are parallel)

CQ is a line segment that is obtained when line segment BC is produced.

∴ AD || CQ

We have,

AD = CQ and AD|| CQ

Hence, ACQD is a parallelogram.

Consider ∆DCQ and ∆ACQ

These are on the same base CQ and between the same parallels CQ and AD.

Therefore,

Area (ΔDCQ) = Area (ΔACQ)

∴ Area (ΔDCQ) - Area (ΔPQC) = Area (ΔACQ) - Area (ΔPQC) [Subtracting Area (ΔPQC) on both sides.]

∴ Area (ΔDPQ) = Area (ΔAPC) ...(2)

From equations (1) and (2), we obtain

Area (ΔBPC) = Area (ΔDPQ), proved.

**Video Solution:**

## In Fig. 9.32, ABCD is a parallelogram and BC is produced to a point Q such that AD = CQ. If AQ intersect DC at P, show that ar (BPC) = ar (DPQ). [Hint : Join AC.]

### Maths NCERT Solutions Class 9 - Chapter 9 Exercise 9.4 Question 4:

**Summary:**

If ABCD is parallelogram and BC is produced to a point Q such that AD = CQ and AQ intersect DC at P in the given figure, then ar (ΔBPC) = ar (ΔDPQ).