Solution: □ABCD is a parallelogram.
P is any interior point.
Draw a line \(\overline {XY}\) parallel to AB through P.
Now ΔAPB = 1/2 □AXYB ……………(1)
[∵ ΔAPB, □AXYB lie on the same base AB and between AB//XY]
Also ΔPCD = 1/2 □CDXY ………………… (2)
[ ∵ ΔPCD; □CDXY lie on the same
base CD and between CD//XY]
Adding (1) & (2), we get
Δ APB + ΔPCD = 1/2 □AXYB + 1/2 □CDXY
= 1/2 [□ AXYB + □ CDXY] [from the fig.)
= 1/2 □ABCD
Hence Proved.
