Question 4 (ii)
In the given figure, P is a point in the interior of a parallelogram ABCD. Show that:
(ii) ar(APD) + ar(PBC) = ar(APB) +ar(PCD)

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Solution

Let us draw a line segment MN, passing through point P and parallel to line segment AD. In parallelogram ABCD,
MN|| AD (By construction) ... (6)
ABCD is a parallelogram.
$∴$ AB || DC(Opposite sides of a parallelogram)
$\Rightarrow$ AM || DN... (7)
From equations(6) and(7), we obtain,
MN|| AD and AM || DN
Therefore, quadrilateral AMND is a parallelogram.
It can be observed that $Δ A P D$ and parallelogram AMND are lying on the same base AD and between the same parallel lines AD and MN.
$∴ A r e a (Δ A P D) = \frac{1}{2} A r e a (A M N D) . . . (8)$
Similarly, for $Δ P C B$ and parallelogram MNCB,
$A r e a (Δ P C B) = \frac{1}{2} A r e a (M N C B) . . . (9)$
Adding equations (8) and (9), we obtain,
$A r e a (Δ A P D) + A r e a (Δ P C B) = \frac{1}{2} [A r e a (A M N D) + A r e a (M N C B)]$
$A r e a (Δ A P D) + A r e a (Δ P C B) = \frac{1}{2} A r e a (A B C D) . . . (10)$
On comparing equations (5) in the previous question and (10), we obtain,
$A r e a (Δ A P D) + A r e a (Δ P B C) = A r e a (Δ A P B) + A r e a (Δ P C D)$

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Similar questions

In Fig, P is a point in the interior of a parallelogram ABCD. Show that
(i) ar(APB)+ar(PCD)=½ ar(ABCD)
(ii) ar(APD)+ar(PBC)=ar(APB)+ar(PCD)

P

A B C D

a r (△ A P D) + a r (△ P B C) = a r (△ A P B) + a r (△ P C D)

a r (A P B) + a r (P C D) = \frac{1}{2} a r (A B C D)

a r (A P D) + a r (P B C) = a r (A P B) + a r (P C D)

In the given figure, P is a point in the interior of a parallelogram ABCD. Show that

(i) ar (APB) + ar (PCD) = ar (ABCD)

(ii) ar (APD) + ar (PBC) = ar (APB) + ar (PCD)

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

(i) a r (A P B) + a r (P C D) = \frac{1}{2} a r (A B C D)

(ii) a r (A P D) + a r (P B C) = a r (A P B) + a r (P C D)

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