Question

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 triangles ABC is divided into four congruent triangles.

Answer 1

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, AD = BD = ${\displaystyle \frac{1}{2}}$AB, BE = EC = ${\displaystyle \frac{1}{2}}$BC

And AF = CF = ${\displaystyle \frac{1}{2}}$AC

Now, using the mid-point theorem,

EF || AB and EF = ${\displaystyle \frac{1}{2}}$AB = AD = BD

ED || AC and ED = ${\displaystyle \frac{1}{2}}$AC = AF = CF

And DF || BC and DF = ${\displaystyle \frac{1}{2}}$BC = BE = CE

In ΔADF and ΔEFD,

AD = EF

AF = DE

And DF = FD   ...[Common]

∴ ΔADF ≅ ΔEFD   ...[By SSS congruence rule]

Similarly, ΔDEF ≅ ΔEDB

And ΔDEF ≅ ΔCFE

So, ΔABC is divided into four congruent triangles.

Hence proved.