# In ∆ ABC, D is the mid-point of AB and P is any point on BC. If CQ || PD meets AB in Q (Fig. 9.14), then prove that ar (BPQ) = 1/2 ar (ABC).

**Solution:**

Given, ABC is a triangle

D is the midpoint of AB

P is any point on BC

CQ || PD meets AB in Q

We have to prove that ar(BPQ) = 1/2 ar(ABC)

Join PQ and CD

Since D is the midpoint of AB. CD is the median of the triangle ABC

We know that the median of a triangle divides it into two triangles of equal areas.

So, ar(BCD) = 1/2 ar(ABC)

From the figure,

ar(BCD) = ar(BPD) + ar(DPC)

Now, ar(BPD) + ar(DPC) = 1/2 ar(ABC) ---------------------- (1)

We know that the area of triangles on the same base and between the same parallel lines are equal.

Triangles DPQ and DPC are on the same base DP and between the same parallel lines DP and CQ.

So, ar(DPQ) = ar(DPC) ----------------------- (2)

Substituting (2) in (1),

ar(BPD) + ar(DPQ) = 1/2 ar(ABC)

From the figure,

ar(BPQ) = ar(BPD) + ar(DPQ)

Therefore, ar(BPQ) = 1/2 ar(ABC)

**✦ Try This: **D, E and F are respectively the mid-points of the sides BC, CA, and AB of a ABC. Show that BDEF is a parallelogram

**☛ Also Check: **NCERT Solutions for Class 9 Maths Chapter 9

**NCERT Exemplar Class 9 Maths Exercise 9.3 Problem 4**

## In ∆ ABC, D is the mid-point of AB and P is any point on BC. If CQ || PD meets AB in Q (Fig. 9.14), then prove that ar (BPQ) = 1/2 ar (ABC)

**Summary:**

In ∆ ABC, D is the mid-point of AB and P is any point on BC. If CQ || PD meets AB in Q (Fig. 9.14), then it is proven that ar (BPQ) = 1/2 ar (ABC)

**☛ Related Questions:**

- ABCD is a square. E and F are respectively the midpoints of BC and CD. If R is the mid-point of EF ( . . . .
- O is any point on the diagonal PR of a parallelogram PQRS (Fig. 9.16). Prove that ar (PSO) = ar (PQO . . . .
- ABCD is a parallelogram in which BC is produced to E such that CE = BC (Fig. 9.17). AE intersects CD . . . .

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