In Figure, P is a point in the interior of a parallelogram ABCD. Show that
(i) ar (\(\triangle{APB}\) + \(\triangle{PCD}\)) = \(\frac{1}{2} \) ar (ABCD)
(ii) ar (\(\triangle{APD}\) + \(\triangle{PBC}\)) = ar(APB) + ar(PCD)

[Hint : Through P, draw a line parallel to AB.]

Construction: Let us draw a line segment EF, passing through point P and parallel to line segment AB.

In parallelogram ABCD, AB || EF ...(i)(By construction)

Also, ABCD is a parallelogram
So, AD || BC
(Opposite sides of parallelogram)
Also, AE || BF ...(ii)
(Since, \(\frac{1}{2} \) AD = AE and \(\frac{1}{2} \) BC = BF)
From equations (i) and (ii), we get,
AB || EF and AE || BF

Therefore, quadrilateral ABFE is a parallelogram.

It can be observed that \(\triangle{APB}\) and parallelogram ABFE are lying on the same base AB and between the same parallel lines AB and EF.

Therefore, we can say that,
Area \(\triangle{APB}\) = \(\frac{1}{2} \) Area (ABFE) ...(iii)
Similarly,
for \(\triangle{PCD}\) and parallelogram EFCD,
\(\Rightarrow \) Area \(\triangle{PCD}\) = \(\frac{1}{2} \) Area (EFCD) ...(iv)
Now, adding eq. (iii) and (iv), we get,
\(\Rightarrow \) Area \(\triangle{APB}\) + Area \(\triangle{PCD}\) = \(\frac{1}{2} \) Area (ABFE) + \(\frac{1}{2} \) Area (EFCD)
\(\Rightarrow \) Area \(\triangle{APB}\) + Area \(\triangle{PCD}\) = \(\frac{1}{2} \)
[Area (ABFE) + Area (EFCD)]
\(\Rightarrow \) Area \(\triangle{APB}\) + Area \(\triangle{PCD}\) = \(\frac{1}{2} \) Area (ABCD) ...(v)

Hence, part i) is proved.



Construction: Let us draw a line segment MN, passing through point P and parallel to line segment AD.

In parallelogram ABCD, MN || AD ...(vi)(By construction)

Also, ABCD is a parallelogram
Thus, AB || DC
(Opposite sides of parallelogram)

Also, AM || DN ...(vii)
(Since, \(\frac{1}{2} \) AB = AM and \(\frac{1}{2} \) DC = DN)
From equations (vi) and (vii), we get,
MN || AD and AM || DN
\(\therefore \) quadrilateral AMND is a parallelogram.


It can be observed that \(\triangle{APD}\) and parallelogram AMND are lying on the same base AD and between the same parallel lines AD and MN.

Therefore, we can say that,
Area \(\triangle{APD}\) = \(\frac{1}{2} \) Area (AMND) ...(viii)

Similarly,
for \(\triangle{PCB}\) and parallelogram MNCB,
\(\Rightarrow \) Area \(\triangle{PCB}\) = \(\frac{1}{2} \) Area (MNCB) ...(ix)

Now, adding eq. (viii) and (ix), we get,

Area \(\triangle{APD}\) + Area \(\triangle{PCB}\) = \(\frac{1}{2} \) Area (AMND) + \(\frac{1}{2} \) Area (MNCB)
\(\Rightarrow \) Area \(\triangle{APD}\) + Area \(\triangle{PCB}\) = \(\frac{1}{2} \) [Area (AMND) + Area (MNCB)]
\(\Rightarrow \) Area \(\triangle{APD}\) + Area \(\triangle{PCB}\) = \(\frac{1}{2} \) Area (ABCD) ...(x)

Thus, comparing eq. (v) and (x), we get,

Area \(\triangle{APB}\) + Area \(\triangle{PCD}\) = Area \(\triangle{APD}\) + Area \(\triangle{PCB}\)

Hence, proved.