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Let $A B C D$ be a parallelogram and $E$ be the mid-point of $A B$. If $P$ is the point of intersection of $D E$ and $A C$, then $\frac{D P}{P E}+\frac{A P}{P C}=$
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The correct answer is:
$\frac{5}{2}$
$A B C D$ is a parallelogram
$E$ is the mid-point of $A B$ $P$ is the intersection of $D E$ and $A C$
In $\triangle P A E$ and $\triangle P C D$,
$\begin{aligned}
& \angle P A E=\angle P C A \Rightarrow \angle A P E=\angle D P C \\
\therefore \quad & \triangle P A E \sim \triangle P C D \\
\because & \frac{P A}{P C}=\frac{P E}{P D}=\frac{A E}{C D} \Rightarrow \frac{P A}{P C}=\frac{P E}{P D}=\frac{A E}{2 A E}\left[A E=\frac{1}{2} C D\right] \\
& \frac{P A}{P C}=\frac{1}{2} \text { and } \frac{P E}{P D}=\frac{1}{2} \quad \therefore \frac{P D}{P E}+\frac{P A}{P C}=2+\frac{1}{2}=\frac{5}{2}
\end{aligned}$
$E$ is the mid-point of $A B$ $P$ is the intersection of $D E$ and $A C$
In $\triangle P A E$ and $\triangle P C D$,
$\begin{aligned}
& \angle P A E=\angle P C A \Rightarrow \angle A P E=\angle D P C \\
\therefore \quad & \triangle P A E \sim \triangle P C D \\
\because & \frac{P A}{P C}=\frac{P E}{P D}=\frac{A E}{C D} \Rightarrow \frac{P A}{P C}=\frac{P E}{P D}=\frac{A E}{2 A E}\left[A E=\frac{1}{2} C D\right] \\
& \frac{P A}{P C}=\frac{1}{2} \text { and } \frac{P E}{P D}=\frac{1}{2} \quad \therefore \frac{P D}{P E}+\frac{P A}{P C}=2+\frac{1}{2}=\frac{5}{2}
\end{aligned}$
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