Question

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

Answer 1

DE // AC and DC // AP

Taking $\Delta$BCD and $\Delta$BPA into consideration;

$\angle$B =$\angle$B (common angle)

$\angle$BDC = $\angle$BAP (corresponding angles)

$\angle$BCD = $\angle$BPA (corresponding angles)

So, $\Delta$BCD ~$\Delta$BPA by AAA criteria

$\frac{BC}{BP}$ = $\frac{BD}{BA}$ ------(i)

Taking $\Delta$BDE and $\Delta$BAC into consideration;

$\angle$B = $\angle$B (common angle)

$\angle$BDE = $\angle$BAC (corresponding angles)

$\angle$BED = $\angle$BCA (corresponding angles)

So, $\Delta$BDE ~$\Delta$BAC by AAA criteria

$\frac{BE}{BC}$ = $\frac{BD}{BA}$ ------(ii)

So from eq.(i) and (ii);

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