So segment AE joins the centers of the circles.
C is the midpoint of AE as given and A,B,C,D,E are
on segment AE (obviously) with B and D are the
tangent points of the circles.
Since FG is perpendicular to AE through C, there are
2 right triangles formed: ACF and ECF.
These right triangles are congruent by LL, as CF is congruent to itself
by reflexive property
Then AF and EF are congruent, CPCTC
which are the radii of the circle.
Since the radii are the same, the circles are congruent.