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

In \triangle ABC, D is the mid-point of AB and P is any point on BC. If CQ\parallel PD meets AB in Q (Fig.), then prove that ar (BPQ)=\frac{1}{2}ar(ABC).

Answers (1)

Solution.
In \triangle ABC, D is the mid-point of AB and P is any point on BC.
CQ\parallel PD
Join D and C

Consider \triangle ABC,
Since D is the mid-point of AB
So CD is the median of \triangle ABC
Using mid-point theorem,
ar(\triangle BCD)=\frac{1}{2}ar(\triangle ABC)          …(i)
Now, CQ \parallel PD
Since \triangle PDQ and \triangle PDC are on the same base PD and between the same parallel lines PD & QC. So we have:
\therefore ar(\triangle PDQ)=ar(\triangle PDC)       …(ii)
From (i) & (ii)
ar (\triangle BCD) = \frac{1}{2}ar(\triangle ABC)
ar(\triangle BPD)+ar(\triangle PDC)=\frac{1}{2}ar(\triangle ABC)
Now, area of \triangle PDC = area of \triangle PDQ
ar(\triangle BPD)+ar(\triangle PDQ)=\frac{1}{2}ar(\triangle ABC)
\Rightarrow ar(\triangle BPQ)=\frac{1}{2}ar(\triangle ABC)
Hence proved.

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