In the given figure, P is a point in the interior of a parallelogram ABCD. Show that
(i) ar (APB) + ar (PCD) = 1/2ar (ABCD)
(ii) ar (APD) + ar (PBC) = ar (APB) + ar (PCD)
[Hint: Through. P, draw a line parallel to AB]
Solution
(i) Let us draw a line segment EF, passing through point P and parallel to line segment AB.
In parallelogram ABCD,
AB || EF (By construction) ... (1)
ABCD is a parallelogram.
∴ AD || BC (Opposite sides of a parallelogram)
⇒ AE || BF ... (2)
From equations (1) and (2), we obtain
AB || EF and AE || BF
Therefore, quadrilateral ABFE is a parallelogram.
It can be observed that ΔAPB and parallelogram ABFE are lying on the same base AB and between the same parallel lines AB and EF.
∴ Area (ΔAPB) = 1/2Area (ABFE) ... (3)
Similarly, for ΔPCD and parallelogram EFCD,
Area (ΔPCD) = 1/2Area (EFCD) ... (4)
Adding equations (3) and (4), we obtain
Area (ΔAPB) + Area (ΔPCD) = 1/2[Area (ABFE) + Area (EFCD)]
Area (ΔAPB) + Area (ΔPCD) = 1/2Area (ABCD) ........(5)
(ii)
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)
⇒ 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 ΔAPD and parallelogram AMND are lying on the same base AD and between the same parallel lines AD and MN.
∴ Area (ΔAPD) = 1/2Area (AMND) ... (8)
Similarly, for ΔPCB and parallelogram MNCB,
Area (ΔPCB) = 1/2Area (MNCB) ... (9)
Adding equations (8) and (9), we obtain
Area (ΔAPD) + Area (ΔPCB) = 1/2[Area (AMND) + Area (MNCB)]
Area (ΔAPD) + Area (ΔPCB) = 1/2Area (ABCD) ...........(10)
On comparing equations (5) and (10), we obtain
Area (ΔAPD) + Area (ΔPBC) = Area (ΔAPB) + Area (ΔPCD)
