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# 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
(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.