In $\Delta ABC$,$D$,$E$,and $F$ are respectively the mid-points of sides $AB$,$BC$,and $CA$. Show that $\Delta ABC$ is divided into four congruent triangles by joining $D$,$E$,and $F$.

(N/A) Since $D$ and $E$ are the mid-points of sides $AB$ and $BC$ of $\Delta ABC$,by the Mid-point Theorem,$DE \parallel AC$ and $DE = \frac{1}{2} AC = AF$.
Similarly,$DF \parallel BC$ and $EF \parallel AB$.
Therefore,$AFDE$,$BDFE$,and $DFCE$ are all parallelograms.
Now,$DE$ is a diagonal of the parallelogram $BDFE$,which divides it into two congruent triangles,so $\Delta BDE \cong \Delta FED$.
Similarly,$DF$ is a diagonal of the parallelogram $AFDE$,so $\Delta DAF \cong \Delta FED$.
Also,$EF$ is a diagonal of the parallelogram $DFCE$,so $\Delta EFC \cong \Delta FED$.
Since all three triangles are congruent to $\Delta FED$,it follows that all four triangles $(\Delta BDE, \Delta DAF, \Delta EFC, \text{ and } \Delta FED)$ are congruent to each other.