In ∆ABC, D is the midpoint of AB. P is any point on BC, and Q is a point on AB such that CQ || PD. Class 9

In ∆ABC, D is the midpoint of AB. P is any point on BC, and Q is a point on AB such that CQ || PD. Class 9

Question 1.

In ∆ABC, D is the midpoint of AB. P is any point on BC, and Q is a point on AB such that CQ || PD. PQ is joined (Fig. 6.34). Prove that Area (∆BPQ) = $\frac{1}{2}$ Area (∆ABC). Class 9

Solution:

Given: D is midpoint of AB. Q is a point on

AB such that CQ || PD

To prove: A(∆BPQ) = $\frac{1}{2}$ A(∆ABC)

Construction: Joint point C and D.

Proof:

D is midpoint of AB. ...[Given]

∴ CD is median of ∆ABC

∴ A(∆BCD) = $\frac{1}{2}$ × A(∆ABC) ...(i) [Median of a triangle divides it into two triangles of equal area.]

But

A(∆BCD) = A(∆BDP) + A(∆PDC) ...(ii)

Now,

CQ || PD ...[Given]

∴ ∆PDC and ∆PDQ lie on same base PD and same parallel lines.

∴ A(∆PDC) = A(∆PDQ) ... (iii)

From (ii),

A(∆BCD) = A(∆BDP) + A(∆PDQ)

∴ A(∆BCD) = A(∆BPQ) .. .(iv)

Substituting (iv) in (i) we get

A (∆BPQ) = $\frac{1}{2}$ A(∆ABC)