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In Delta ABC, D is the mid-point of AB a...

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)`.

Text Solution

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GIVEN:
`D` is the midpoint of side `AB` of `triangle ABC` and `P` is any point on `BC`. `CQ||PD` meets `AB` in `Q`.
TO PROVE:
`ar(triangleBPQ)=(1)/(2)ar(triangleABC)`.
CONSTRUCTION: Join `CD` and `PQ`.
PROOF:
We know that a median of a triangle divides it into two triangles of equal area.
And, in `triangleABC`, `CD` is a median.
`therefore ar(triangleBCD)=(1)/(2)ar(triangleABC)`
`rArr ar(triangleBPD)+ar(triangleDPC)=(1)/(2)ar(triangleABC)" "...(i)`
But, `triangleDPC` and `DPQ` being on the same base `DP` and between the same parallels `DP` and `CQ`, we have
`ar(triangleDPC)=ar(triangleDPQ)" "...(ii)`
Using (ii) in(i), we get
`ar(triangleBPQ)+ar(triangleDPQ)=(1)/(2)ar(triangleABC)`
`therefore ar(triangleBPQ)=(1)/(2)ar(triangleABC).`
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