In `Delta` ABC, D is the mid-point of AB and P is any point on BC. If – InterviewSolution

InterviewSolution

1.	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)`.
Answer» 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 `triangle`ABC, CD is a median. `therefore ar(triangleBCD)=(1)/(2)ar(triangleABC)` `rArr ar(triangleBPD)+ar(triangleDPC)=(1)/(2)ar(triangleABC)" "...(i)` But, `triangle`DPC 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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