In a parallelogram $ABCD$,$E$ and $F$ are the mid-points of sides $AB$ and $CD$ respectively (see Fig). Show that the line segments $AF$ and $EC$ trisect the diagonal $BD$.

(N/A) Given: $ABCD$ is a parallelogram where $E$ and $F$ are mid-points of $AB$ and $CD$ respectively.
To prove: $AF$ and $EC$ trisect the diagonal $BD$,i.e.,$DP = PQ = QB$.
Proof:
Since $ABCD$ is a parallelogram,$AB \parallel DC$ and $AB = DC$.
Since $E$ and $F$ are mid-points,$AE = \frac{1}{2}AB$ and $FC = \frac{1}{2}DC$.
Since $AB = DC$,we have $AE = FC$.
Also,$AE \parallel FC$ because $AB \parallel DC$.
Thus,$AECF$ is a parallelogram (a quadrilateral with one pair of opposite sides equal and parallel).
Therefore,$AF \parallel EC$.
In $\Delta DQC$,$F$ is the mid-point of $DC$ and $FP \parallel CQ$ (since $AF \parallel EC$). By the converse of the mid-point theorem,$P$ is the mid-point of $DQ$,so $DP = PQ$ ... $(1)$.
In $\Delta ABP$,$E$ is the mid-point of $AB$ and $EQ \parallel AP$ (since $AF \parallel EC$). By the converse of the mid-point theorem,$Q$ is the mid-point of $BP$,so $PQ = QB$ ... $(2)$.
From $(1)$ and $(2)$,$DP = PQ = QB$.
Hence,$AF$ and $EC$ trisect the diagonal $BD$.