In the adjoining figure, ABCD is a parallelogram and P is any points o – InterviewSolution

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1.	In the adjoining figure, ABCD is a parallelogram and P is any points on BC. Prove that `ar(triangleABP)+ar(triangleDPC)=ar(trianglePDA)`.
Answer» Draw `ALbotBC and PMbotAD`. Since BC\|\|AD, so distance between them remain the same. `therefore AL=PM`. `=(1)/(2)xxBPxxAL+(1)/(2)xxPCxxAL=(1)/(2)xxALxx(BP+PC)` `=(1)/(2)xxALxxBC=(1)/(2)xxPMxxAD" "[therefore AL=PM and BC = AD]` `=ar(triangle PDA)`.

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