Question

A parallelogram $ABCD$ has $E$ as the midpoint of $DC$ and $F$ a point on $AC$ such that $CF=\frac{1}{4}AC$. EF produced meets $BC$ to $G$. Prove that $G$ is the midpoint of $BC$.

Answer 1

Given $E$ is midpoint of $DC$ and $CF=\frac{1}{4}AC$

$OC=\frac{1}{2}AC................................\left(1\right)$

$CF=\frac{1}{4}AC................\left(2\right)$

Comparing $1$ and $2$ we get $CF=\frac{1}{2}OC$

In $△DCO$, $E$ and $F$ are mid points of the sides $DC$ and $OC$ respectively.

Hence, $EF||OD$ ...... (By mid point theorem)

In $△COB$, $FG$ is parallel to $OB$ (since $DOB$ is on same line $DO||EG||OB$) and $F$ is mid point of one side

Hence $G$ will be the mid point of other side of the triangle where the line $EG$ touches $BC$(By converse of mid point theorem)

Hence proved.