Question

In the adjoining figure, medians $AD$ and $CE$ of $△ABC$ meet at the point $G$, and $DF$ is drawn parallel to $CE$. Prove that
(i) $EF=FB$
(ii) $AG:GD=2:1$.

Answer 1

In the following fig, $AD$ and $CE$ are medians of $\Delta ABC,$ $DF$ is drawn parallel to $CE$

(1) $EF=FB$

In $\Delta BFD$ and $\Delta BEC$

$\angle BFD=\angle BEC$ (Corresponding angles)

$\angle FBD=\angle EBC$ (Common angles)

$\Delta BFD\sim \Delta BEC$ (AA similarity)

$\therefore \frac{BF}{BE}=\frac{BD}{BC}$

$\therefore \frac{BF}{BE}=\frac{1}{2}$ (As $D$ is the mid point of $BC$)

$BE=2BF$

$BF+FE=2BF$

Hence $EF=FB$

ii) $AG:GD=2:1$

In $\Delta AFD,EG||FD$.

Then using proportionality theorem,

$\therefore \frac{AE}{EF}=\frac{AG}{GD}$........(1)

Now $AE=EB$ (as $E$ is the mid of $AB$)

$AE=2EF$ (Since $EF=FB$ by $1$)

$AG:GD=2:1$..........From 1

Hence $AG:GD=2:1$