Question

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

Answer 1

In $△BPA$, we have

$DC||AP$ [Given]

Therefore, by basic proportionality theorem, we have

$\frac{BC}{CP}=\frac{BD}{DA}$

In $△BCA$, we have

$DE||AC$

Therefore, by basic proportionality theorem, we have

$\frac{BE}{EC}=\frac{BD}{DA}$

From (i) and (ii), we get

$\frac{BC}{CP}=\frac{BE}{EC}$ or $\frac{BE}{EC}=\frac{BC}{CP}$ $\left[Henceproved\right]$