$D, E$ and $F$ are respectively the mid-points of the sides $AB, BC$ and $CA$ of a triangle $ABC$. Prove that by joining these mid-points $D, E$ and $F$,the triangle $ABC$ is divided into four congruent triangles.

(N/A) Given: $A$ $\triangle ABC$ and $\triangle DEF$ which is formed by joining the mid-points $D, E$ and $F$ of the sides $AB, BC$ and $CA$ of $\triangle ABC$.
To prove: $\triangle ADF \cong \triangle DBE \cong \triangle ECF \cong \triangle DEF$
Proof: $D$ and $F$ are the mid-points of $AB$ and $AC$ respectively.
By the Mid-point Theorem,$DF \parallel BC$ and $DF = \frac{1}{2} BC = BE = EC$.
Similarly,$DE \parallel AC$ and $DE = \frac{1}{2} AC = AF = FC$.
Also,$EF \parallel AB$ and $EF = \frac{1}{2} AB = AD = DB$.
In quadrilateral $ADFE$,$AD \parallel EF$ and $AF \parallel DE$,so it is a parallelogram. Thus,$\triangle ADF \cong \triangle FED$ (diagonal $DF$ divides parallelogram into two congruent triangles).
Similarly,$BDEF$ is a parallelogram,so $\triangle DBE \cong \triangle FED$.
And $CEFD$ is a parallelogram,so $\triangle ECF \cong \triangle FED$.
Therefore,$\triangle ADF \cong \triangle DBE \cong \triangle ECF \cong \triangle DEF$.
Hence,the triangle $ABC$ is divided into four congruent triangles.