p is a point in the interiarof a parallogram ABCD
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P is a point in the interior of a parallelogram ABCD. Show that ar(△APD)+ar(△PBC)=ar(△APB)+ar(△PCD)
ABCD is a ∥gm such that AB∥CD & AD∥BC. Draw line GM passing through P, parallel to AD & BC. i.e. GH∥AD∥BC.
Here, GH∥AD (By construction) & AG∥DH (∵ AB∥CD & G & H are points on AB & CD respectively).
∴ AGHD is a ∥gm.
Now, △APD and ∥gm AGHD are on the same base AD and between the bank ∥als AD & Gm.
Therefore,
ar △APD =
2
1
ar (AGHD) ⟶(1)
Similarly ar (△PCB) =
2
1
ar (GBCH) ⟶(2)
Adding equations (1) & (2), we get
ar △APD + ar △PCB =
2
1
(ar AGHD + ar GBCH)
⇒ ar △APD + ar △PCB =
2
1
ar ABCD
Also,
2
1
ar ABCD = ar △APB + ar △PCD
∴ ar △APD + ar △PBC = ar △APB + ar △PCD
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