Question

In a $△ABC$ ,$D,E$ and $F$ are respectively mid points of sides $AB.BC$ and $CA$, Show that $△ABC$ is divided in to four congruent triangles by joining $D,E$ and $F$.

Answer 1

R.E.F image

to prove:

$\Delta ADF\cong \Delta BDE\cong \Delta DEF\cong \Delta EFC$

Proof.

$\because D,E,F$ is the mid point of

$AB,BC,AC$

$\therefore \Delta ADF\cong \Delta ABC$ (by $\frac{AD}{BD}=\frac{AF}{FC}$ mean $DF\parallel BC$ )

Hence $DF=\frac{1}{2}BC=BE=EC$

Similarly $DE=\frac{1}{2}AC=AF=FC$

$FC=\frac{1}{2}AB=AD=BD$...(1)

In $\Delta ADF$ & $\Delta BDE$

$AD=BD$ (given)

$DF=BE$ (by (1))

$AF=DE$ (by (1))

$\therefore \Delta ADF\cong \Delta BDE(S-S-S)$

Similarly

$\Delta BDE\cong \Delta EFC$

$\Delta BDE\cong \Delta DEF$

$\therefore ADF\cong \Delta BDE\cong \Delta DEF\Delta EFC$

Prove.