(N/A) We have: $\triangle ABC$ such that the mid-points of $BC, CA$ and $AB$ are respectively $D, E$ and $F$.
To prove that $BDEF$ is a parallelogram.
In $\triangle ABC$,$E$ and $F$ are the mid-points of $AC$ and $AB$ respectively.
By the Mid-point Theorem,the line segment joining the mid-points of two sides of a triangle is parallel to the third side and is half of it.
Therefore,$EF || BC$ and $EF = \frac{1}{2} BC$.
Since $D$ is the mid-point of $BC$,we have $BD = \frac{1}{2} BC$.
Thus,$EF || BD$ and $EF = BD$.
Since $BDEF$ is a quadrilateral in which one pair of opposite sides ($EF$ and $BD$) is parallel and equal in length,$BDEF$ is a parallelogram.