Fill in the blanks: In the given figure, \( \mathrm{ABCD} \) and \( \mathrm{EFGD} \) are two par...

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Fill in the blanks:
In the given figure, \( \mathrm{ABCD} \) and \( \mathrm{EFGD} \) are two parallelograms and \( \mathrm{G} \) is the mid point of \( \mathrm{CD} \). Then, \( \operatorname{ar}(\triangle \mathrm{DPC}): \operatorname{ar}(\mathrm{EFGD})= \)
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