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


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]
Consider ΔACD.
Line-segment CD is bisected by AB at O. Therefore, AO is the median of
ΔACD.
∴ Area (ΔACO) = Area (ΔADO) ... (1)
Considering ΔBCD, BO is the median.
∴ Area (ΔBCO) = Area (ΔBDO) ... (2)
Adding equations (1) and (2), we obtain
Area (ΔACO) + Area (ΔBCO) = Area (ΔADO) + Area (ΔBDO)
⇒ Area (ΔABC) = Area (ΔABD)



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