Question

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

Answer 1

Given in a $\Delta ABC,D,E$ and F respectively the mid-points of the sides AB, BC and CA.

To prove $\Delta ABC$ is divided into four congruent triangles.

Proof Since, ABC is a triangle and D, E and F are the mid-points of sides AB, BC and CA, respectively
Then, $AD=BD=\frac{1}{2}AB,BE=EC=\frac{1}{2}BC$
And $AF=CF=\frac{1}{2}AC$
Now, using the mid-point theorem,
$EF||ABandEF=\frac{1}{2}AB=AD=BDED||ACandED=\frac{1}{2}AC=AF=CF$
And $DF||BCandDF=\frac{1}{2}BC=BE=CE$
In $\Delta ADFand\Delta EFD,$ $AD=EF$
$AF=DE$
And DF = FD [Common]
$\therefore$ $\Delta ADF\cong \Delta EFD$ [by SSS congruence rule]
Similarly, $\Delta DEF\cong \Delta EDB$
And $\Delta DEF\cong \Delta CFE$
So, $\Delta ABC$ is divided into four congruent triangles. Hence proved.