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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).
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Solution
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.
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