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$ABCD$ is a parallelogram, $P$ is a point on side $BC$ and $DP$ when produced meets $AB$ produced at $L$. Prove that
(i) $\frac{DP}{PL} = \frac{DC}{BL}$
(ii) $\frac{DL}{DP} = \frac{AL}{DC}$
(iii) If $LP : PD = 2 : 3$ then find $BP : BC$
(i) $\frac{DP}{PL} = \frac{DC}{BL}$
(ii) $\frac{DL}{DP} = \frac{AL}{DC}$
(iii) If $LP : PD = 2 : 3$ then find $BP : BC$
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(i) $$\begin{aligned}& \triangle DPC \sim \triangle LPB \\ & \Rightarrow \frac{DP}{PL} = \frac{PC}{PB} = \frac{DC}{BL} \quad \text{(i)} \\ & \text{(ii) As } BC \parallel AD \\ & \therefore \triangle LPB \sim \triangle LDA \\ & \text{In } \triangle DLA, AD \parallel BP \\ & \Rightarrow \frac{LP}{DP} = \frac{LB}{AB} \\ & \Rightarrow \frac{LP}{DP} + 1 = \frac{LB}{AB} + 1 \\ & \Rightarrow \frac{DL}{DP} = \frac{AL}{AB} \\ & \Rightarrow \frac{DL}{DP} = \frac{AL}{CD} \quad (AB = CD) \\ & \text{(iii) } \frac{LP}{LD} = \frac{PB}{AD} \quad (\triangle LPB \sim \triangle LDA) \\ & \Rightarrow \frac{2}{5} = \frac{PB}{BC} \quad (AD = BC)\end{aligned}$$