In parallelogram $ABCD$,$P$ is a point on side $BC$. The line segment $DP$ intersects the extension of side $AB$ at point $L$. Prove that: $(1) \frac{DP}{PL} = \frac{DC}{BL}$ and $(2) \frac{DL}{DP} = \frac{AL}{DC}$.

Given: $ABCD$ is a parallelogram. $P$ is on $BC$,and $DP$ produced meets $AB$ produced at $L$.
$(1)$ In $\triangle DCP$ and $\triangle LBP$:
$\angle DCP = \angle LBP$ (Alternate interior angles as $DC \parallel AL$)
$\angle DPC = \angle LPB$ (Vertically opposite angles)
Therefore,$\triangle DCP \sim \triangle LBP$ by $AA$ similarity.
Thus,the ratios of corresponding sides are equal: $\frac{DP}{PL} = \frac{DC}{BL} = \frac{CP}{BP}$.
Hence,$\frac{DP}{PL} = \frac{DC}{BL}$ is proved.
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