Question

In Fig, P is a point in the interior of a parallelogram ABCD. Show that
(i) ar(APB)+ar(PCD)=½ ar(ABCD)
(ii) ar(APD)+ar(PBC)=ar(APB)+ar(PCD)

Answer 1

Since ABCD is a parallelogram AB∥CD and AD∥BC

(opposite side of parallelogram are parallel)

We draw line EF passing through P, parallel to AB and DC

i.e., EF∥CD

Here, AE∥BF (As AD∥BC and parts of parallel AB∥EF sides are parallel )

∴EFBA is a parallelogram

Similarly EFCD is a parallelogram

Now △APB and parallelogram EFBA are on same base AB and between the same parallel lines AB and EF

∴ area (△APB)=12 area (EFBA)−−−−−−−−−−(1)

Similarly

△PCD and parallelogram EFCD are on same base CD and between the same parallel lines CD and EF ∴ area (△PCD)=½ area (EFCD)−−−−−−−−−(2)

adding (1) and (2) , we obtain

ar(△APB)+ar(△PCD)=½ ar(EFAB)+½ ar(EFCD)

ar(△APB)+ar(△PCD)=½ ar(ABCD)