Question

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

Answer 1

Given: ABC is a triangle and D, E and F are mid-points of sides AB, BC and CA, respectively.

To prove: ∆ABC is divided into four congruent triangles.

Construction: Join DE, EF and FD.

Proof: Given 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, by mid-point theorem,

EF || AB

and EF = \(\frac { 1 }{ 2 }\) AB = AD = BD

ED || AC

and ED = \(\frac { 1 }{ 2 }\) AC = AF = CF

DF || BC

and DF = \(\frac { 1 }{ 2 }\) BC = BE = EC.

Now in ∆ADF and ∆EFD,

AD = EF (opposite sides of a parallelogram)

DF = FD (common sides)

and DE = AF

∆ADF = ∆EFD (by SSS congruence rule)

Similarly, ∆DEF = ∆DEB and ∆DEF = ∆CEF

Thus, ∆ABC is divided into four congruent triangles.

Hence proved.