Question

If three concurrent straight lines $AD,BE$ and $CF$ are drawn from the angular points of a $△ABC$ such that $\frac{AF}{FB}=\frac{1}{2},\frac{BD}{DC}=\frac{2}{3},$ then $\frac{EA}{EC}$ is equal to

• A. $1:3$
• B. $2:3$
• C. $3:2$
• D. $1$

Answer 1

The correct option is A $1:3$

If $AD,BE,CF$ are concurrent lines meeting at the point, (let say) $O$ of a $△ABC$ then
$\frac{AF}{FB}\times \frac{BD}{DC}\times \frac{EC}{EA}=1$
$\Rightarrow \frac{1}{2}\times \frac{2}{3}\times \frac{EC}{EA}=1$
$\therefore \frac{EA}{EC}=\frac{1}{3}$