Question

D, E, and F are respectively the mid-points of the sides AB, BC and CA of a $\Delta ABC$. Triangles are formed by joining these mid-points D, E and F. Which among the following is true?

• A. $\Delta EFD\cong \Delta CFE$
• B. $\Delta DEF\cong \Delta EDB$
• C. $\Delta ADF\cong \Delta EFD$
• D. $Alltheabove$

Answer 1

The correct option is D $Alltheabove$

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

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, $\Delta EFD\cong \Delta CFE$
So,
$\Delta ABC$ is divided into four congruent triangles.
Hence proved.