Question

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

Answer 1

Given: In ∆ABC, D is the mid-point of AB and P is any point on BC.

CQ || PD means AB in Q.

To prove: ar (∆BPQ) = `1/2` ar (∆ABC)

Construction: Join PQ and CD.

Proof: Since, D is the mid-point of AB. So, CD is the median of ∆ABC.

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

∴ ar (∆BCD) = `1/2` ar (∆ABC)

⇒ ar (∆BPD) + ar (∆DPC) = `1/2` ar (∆ABC)  ...(i)

Now, ∆DPQ and ∆DPC are on the same base DP and between the same parallel lines DP and CQ.

So, ar (∆DPQ) = ar (∆DPC)  ...(ii)

On putting the value from equation (ii) in equation (i), we get

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

⇒ ar (∆BPQ) = `1/2` ar (∆ABC)

Hence proved.