Let ABCD be a parallelogram and E be the mid- | vedclass

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(A) Given that $ABCD$ is a parallelogram and $E$ is the mid-point of $AB$. Since $AB \parallel CD$,we have $AE \parallel CD$. In $	riangle PAE$ and $

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$\frac{5}{2}$

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(A) Given that $ABCD$ is a parallelogram and $E$ is the mid-point of $AB$. Since $AB \parallel CD$,we have $AE \parallel CD$. In $	riangle PAE$ and $	riangle PCD$,$\angle PAE = \angle PCD$ (alternate interior angles) $\angle APE = \angle CPD$ (vertically opposite angles) Therefore,$	riangle PAE \sim 	riangle PCD$ by $AA$ similarity. From the property of similar triangles,we have: $\frac{PA}{PC} = \frac{PE}{PD} = \frac{AE}{CD}$. Since $E$ is the mid-point of $AB$,$AE = \frac{1}{2} AB$. Also,$AB = CD$ in a parallelogram,so $AE = \frac{1}{2} CD$. Thus,$\frac{PA}{PC} = \frac{PE}{PD} = \frac{1}{2}$. This implies $\frac{PA}{PC} = \frac{1}{2}$ and $\frac{PD}{PE} = 2$. Therefore,$\frac{DP}{PE} + \frac{AP}{PC} = 2 + \frac{1}{2} = \frac{5}{2}$.

Let $ABCD$ be a parallelogram and $E$ be the mid-point of $AB$. If $P$ is the point of intersection of $DE$ and $AC$,then $\frac{DP}{PE} + \frac{AP}{PC} = $

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