Question

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

Answer 1

In $△$BPA, we have

DC $\parallel$ AP [Given]

$\therefore$ by basic prportionality theorem, we have

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

In $△$BCA, we have

DE $\parallel$AC [Given]

$\therefore$ by basic proportionality theorem, we have

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

From (i) and (ii), we get

$\frac{BC}{CP}$ = $\frac{BE}{BC}or\frac{BE}{BC}=\frac{BC}{CP}$