$AD, BE$ and $CF$ are three concurrent lines in $\triangle ABC$, meeting the opposite sides in $D, E$ and $F$ respectively. Show that the joins of the midpoints of $BC, CA$ and $AB$ to the midpoints of $AD, BE$ and $CF$ are concurrent.
Let $D', E'$ and $F'$ be the midpoints of $BC, CA$ and $AB$. Then consider $\triangle ABD$. $E'F' \parallel BC$ and therefore in $\triangle ABD$, $E'F'$ cuts $AD$ at its midpoint (By the midpoint theorem). Similiarly $E'D'$ cuts $CF$ at its midpoint and $D'F'$ cuts $BC$ at its midpoint.
How do I proceed with this?