From the figure we know that F and E are the mid points of AB and AC
Based on the mid-point theorem
EF = ½ BC
In the same way
FD = ½ AC and ED = ½ AB
Consider △ AFE and △ BFD
We know that AF = FB
Based on the midpoint theorem
FE = ½ BC = BD
FD = ½ AC = AE
By SSS congruence criterion
△ AFE ≅ △ BFD
Consider △ BFD and △ FED
We know that FE || BC
So we get FE || BD and AB || ED
Using the midpoint theorem
FB || ED
Hence, BDEF is a parallelogram
So we know that FD is a diagonal which divides the parallelogram into two congruent triangles
△ BFD ≅ △ FED
In the same way we can prove that FECD is a parallelogram
△ FED ≅ △ EDC
So we know that △ BFD, △ FDE, △ FED and △ EDC are congruent to each other.
Therefore, it is proved that the line segments joining the middle points of the sides of a triangle divide it into four congruent triangles.
