In parallelogram ABCD,A-P-B and AP = frac{2}{ | vedclass

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(C) In parallelogram $ABCD$,$AB \parallel CD$ and $AB = CD$. Given $AP = \frac{2}{3} AB$,we have $AP = \frac{2}{3} CD$. Since $AB \parallel CD$,$\Delt

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$4:9$

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(C) In parallelogram $ABCD$,$AB \parallel CD$ and $AB = CD$. Given $AP = \frac{2}{3} AB$,we have $AP = \frac{2}{3} CD$. Since $AB \parallel CD$,$\Delta APQ \sim \Delta CDQ$ by $AA$ similarity criterion. The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides. Therefore,$\frac{	ext{Area}(\Delta APQ)}{	ext{Area}(\Delta CDQ)} = \left( \frac{AP}{CD} ight)^2$. Substituting $AP = \frac{2}{3} CD$,we get $\frac{	ext{Area}(\Delta APQ)}{	ext{Area}(\Delta CDQ)} = \left( \frac{2/3 CD}{CD} ight)^2 = \left( \frac{2}{3} ight)^2 = \frac{4}{9}$.

In parallelogram $ABCD$,$A-P-B$ and $AP = \frac{2}{3} AB$. $\overline{DP}$ intersects $\overline{AC}$ at $Q$. Find the ratio of the areas of $\Delta APQ$ and $\Delta CDQ$.

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