Question

In the given figure $DE\parallel AC$ and $DF\parallel AE$. Prove that.
$\frac{BF}{FE}=\frac{BE}{EC}$

Answer 1

Given: $D$ is any point on side $AB$ of $△ABC$ and $E$ and $F$ are two points on side $BC$, line segment $DF,DE$ and $AE$ are drawn

To prove that: $\frac{BF}{FE}=\frac{BE}{EC}$

Proof: In $△BCA$

$DE\parallel AC$ (given)
By basic proportionality theorem, if two triangles are similar then their corresponding sides are proportional.

$\frac{BE}{EC}=\frac{BD}{DA}$...(i)

Again in $△BEA$, $DF\parallel AE$ (given)

$\frac{BF}{FE}=\frac{BD}{DA}$....(ii) (by basic proportionality theorem)

From equation (i) and (ii), we can write

$\frac{BF}{FE}=\frac{BE}{EC}$