Question

$D,E$ and $F$ are the mid-points of the sides $AB,BC$ and $CA$ respectively of $△ABC$. $AE$ meets $DF$ at $O.P$ and $Q$ are the mid-points of $OB$ and $OC$ respectivley. Prove that $DPQF$ is a parrallelogram.

Answer 1

In $△ABC,D$ is the midpoint of $AB$ and $F$ is the midpoint of $CA$.

Hence by midpoint theorem, $DF\parallel BC$ and $DF=\frac{1}{2}BC$ ---(1)

In $△BOC,P$ is the midpoint of $OB$ and $Q$ is the midpoint of $OC$.

By midpoint theorem, $PQ\parallel BC$ and $PQ=\frac{1}{2}BC$ ---(2)

In $△BOA,P$ is the midpoint of $OB$ and $D$ is the midpoint of $BA$.

By midpoint theorem, $PD\parallel OA$ and $PD=\frac{1}{2}OA$ ---(3)

In $△COA,Q$ is the midpoint of $OC$ and $F$ is the midpoint of $CA$.

By midpoint theorem, $FQ\parallel OA$ and $FQ=\frac{1}{2}OA$ ---(4)

From (1) and (2), $DF\parallel PQ$ and $DF=PQ$

From (3) and (4), $PD\parallel QF$ and $PD=QF$

$\therefore DPQF$ is a parallelogram.