In the given figure, $DE \parallel AC$ and $DC \parallel AP$. Prove that $\frac{BE}{EC} = \frac{BC}{CP}$.

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Question

In the given figure, $DE \parallel AC$ and $DC \parallel AP$. Prove that $\frac{BE}{EC} = \frac{BC}{CP}$.

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Accepted Answer

Step 1: In $	riangle ABC$, $DE \parallel AC$.
By BPT: $\frac{BD}{DA} = \frac{BE}{EC}$ ... (i)
Step 2: In $	riangle ABP$, $DC \parallel AP$.
By BPT: $\frac{BD}{DA} = \frac{BC}{CP}$ ... (ii)
From (i) and (ii): $\frac{BE}{EC} = \frac{BC}{CP}$

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