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

Explanation

Given In `Delta` ABC, D is the mid-point of AB and P is any point on BC.
`CQ || PD` means AB in Q.
To prove `" " ar(DeltaBPQ) = (1)/(2) ar (DeltaABC)`
Construction Join PQ and CD.
Proof Since, D is the mid - point of AB. So CD is the median of `DeltaABC`.
We know that, a median of a triangle divides it into two triangles of equal areas.

`therefore" "` `ar (DeltaBCD) = (1)/(2) ar (DeltaABC)`
`rArr` `" " ar (DeltaBPD) + ar (DeltaDPC) = (1)/(2) ar (DeltaABC)" "` ...(i)
Now, `DeltaDPQ` and `DeltaDPC` are on the same base DP and between the same parallel lines DP andCQ.
So, `" " ar (DeltaDPQ) = ar (DeltaDPC)" "` ...(ii)
On putting the value from Eq. (ii) in Eq. (i), we get
`ar (DeltaBPD) + ar (DeltaDPQ) = (1)/(2)ar(DeltaABC)`
`rArr" "` `ar (DeltaBPQ) = (1)/(2) ar (DeltaABC)" "` Hence proved.