Question

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

Answer 1

In $△$s BDC and BAP,
$\angle DBC=\angle PBA$ (Common angle)
$\angle BDC=\angle BPA$ (Corresponding angles)
$\angle BCD=\angle BPA$ (Corresponding angles)
Thus, $△BDC\sim △BAP$ (AAA rule)
Hence, $\frac{BC}{BP}=\frac{BD}{AB}$ ..... $\left(i\right)$ (Corresponding sides)

Similarly, $△BDE\sim △BAC$

Hence, $\frac{BD}{AB}=\frac{BE}{BC}$ ...... $\left(ii\right)$ (Corresponding sides)

Thus from $\left(i\right)$ and $\left(ii\right)$, we get

$\frac{BE}{BC}=\frac{BC}{BP}$

$\frac{BC}{BE}=\frac{BP}{BC}$ ....... (Reciprocating)

$\frac{BC}{BE}-1=\frac{BP}{BC}-1$

$\frac{BC-BE}{BE}=\frac{BP-BC}{BC}$

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

$\frac{BE}{EC}=\frac{BC}{CP}$ ...... (Reciprocating)