D, E and F are respectively the mid-points of the sides AB, BC and CA of a ΔABC. Prove that by joining these mid-points D, E and F, the ΔABC is
divided into four congruent triangles.
Given In a ΔABC, D, E and F are respectively the mid-points of the sides AB, BC and CA. To prove ΔABC 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, $\quad A D=B D=\frac{1}{2} A B, B E=E C=\frac{1}{2} B C$
and $A F=C F=\frac{1}{2} A C$
Now, using the mid-point theorem,
$E F \| A B$ and $E F=\frac{1}{2} A B=A D=B D$
$E D \| A C$ and $E D=\frac{1}{2} A C=A F=C F$
and
$A F=C F=\frac{1}{2} A C$
Now, using the mid-point theorem,
$E F \| A B$ and $E F=\frac{1}{2} A B=A D=B D$
$E D \| A C$ and $E D=\frac{1}{2} A C=A F=C F$
and
$D F \| B C$ and $D F=\frac{1}{2} B C=B E=C E$
In $\triangle A D F$ and $\triangle E F D$, $A D=E F$
$A F=D E$
and $D F=F D$ [common]
$\begin{array}{lll}\therefore & \Delta A D F \equiv \Delta E F D & \text { [by SSS congruence rule] }\end{array}$
Similarty, $\triangle D E F \equiv \triangle E D B$
and $\triangle D E F \cong \triangle C F E$
So, $\triangle A B C$ is divided into four congruent triangles.
Hence proved.
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