P is a point in the interior of a parallelogram ABCD. Show that ar (

1.	P is a point in the interior of a parallelogram ABCD. Show that ar (ΔAPB) + ar (ΔPCD) = 1/2 ar(ABCD) (Hint : Through P, draw a line parallel to AB)
Answer» 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.

P is a point in the interior of a parallelogram ABCD. Show that ar (ΔAPB) + ar (ΔPCD) = 1/2 ar(ABCD) (Hint : Through P, draw a line parallel to AB)

Answer»

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.