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In the adjoining figure, ABCD is a paral...

In the adjoining figure, ABCD is a parallelogram and P is any points on BC. Prove that
`ar(triangleABP)+ar(triangleDPC)=ar(trianglePDA)`.

Text Solution

Verified by Experts

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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