Question

P is any point on the side AB of parallelogram ABCD. Prove that area (ΔAPD) + area (ΔPBC) = `1/2` area (|| gm ABCD).

Answer 1

We know that,

If a triangle and parallelogram are drawn on the same base and between the same parallel lines, then the area of the triangle is half of the area of the parallelogram.

Since ΔCPD and || gm ABCD are on the same base DC and between the same || lines AB and DC.

⇒ ar(ΔCPD) = `1/2` ar(|| gm ABCD)  ...(i)

From the figure, 

ar(|| gm ABCD) = ar(ΔAPD) + ar(ΔCPD) + ar(ΔPBC)

Using equation (i), we get

ar(|| gm ABCD) = `1/2` ar(|| gm ABCD) + ar(ΔAPD) + ar(ΔPBC)

⇒ `1/2` ar(|| gm ABCD) = ar(ΔAPD) + ar(ΔPBC)

Hence proved.