Solution :
We know that, the line joining the mid points of two sides of a triangle is parallel to the third side and half of it .
$FE=\large\frac{1}{2}$$BC$
=> $\large\frac{FD}{AC} =\large\frac{1}{2}$
$\large\frac{FE}{BC} = \frac{ED}{AB} =\frac{FD}{AC}$
$\qquad= \large\frac{1}{2}$
If in 2 triangles,side of one triangle is proportional to the sides of another triangle then their corresponding angle are equal .
By similarly $ABC \approx \Delta FDE$