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(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.]

Answer :

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.

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