Question

In ∆ABC, D is the midpoint of AB and P is any point on BC. If CQ || PD meets AB in Q, then prove that $\mathrm{ar}(∆BPQ)=\frac{1}{2}\mathrm{ar}(∆ABC)$.

Answer 1

Given: ∆ABC, D is the midpoint of AB and P is any point on BC.
To prove: ar(∆BPQ) = $\frac{1}{2}$ × ar(∆ABC)​
Construction: Join CD.
Proof: Now, in ∆ABC, CD is a median.
∴ ar(∆BDC) = $\frac{1}{2}$ × ar(∆ABC)
⇒ ar(∆BPD) + ar(∆PDC) = $\frac{1}{2}$ × ar (∆ABC)
∆PDC and ∆PQD are on the same base PD and between the same parallel lines.
Then ar(∆PDC) = ar(∆PQD)

Now, ar(∆BPD) + ar(∆ PQD) = $\frac{1}{2}$ × ar(∆ABC)​
∴ ar ( ∆BPQ) = $\frac{1}{2}$× ar ( ∆ABC)​​ [ ∵​ ar(∆BPQ) = ar(∆BPD) + ar(∆PQD)]
Hence, proved.