In Fig. \( \quad \mathrm{P} \) is a point in the interior of a parallelogram \( \mathrm{ABCD} \)...
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In Fig. \( \quad \mathrm{P} \) is a point in the interior of a parallelogram \( \mathrm{ABCD} \). Show that
(i) \( \operatorname{ar}(\mathrm{APB})+\operatorname{ar}(\mathrm{PCD})=\frac{1}{2} \operatorname{ar}(\mathrm{ABCD}) \)
- (ii) \( \operatorname{ar}(\mathrm{APD})+\operatorname{ar}(\mathrm{PBC})=\operatorname{ar}(\mathrm{APB})+\operatorname{ar}(\mathrm{PCD}) \)
_[Hint : Through P, draw a line parallel to AB.]
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