The drawing below is to scale.  The one given is not.


Let O be the center of the larger semicircle.
Let P be the center of the smaller semicircle.
Draw OP which will pass through the common point of tangency T and will
be perpendicular to tangent EG.
Let the radius of the small semicircle be r = CP = TP = FP
Since AB = 18, the diameter of the larger semicircle, the radius = OB = 9.
OB² + BP² = OP²
9² + (BC-CP)² = (OT+TP)²
9² + (6-r)² = (9+r)²
Solve that and get r = CP = TP = FP = 1.2
Triangles OBP and GTP are similar right triangles for they have
a common acute angle at P.  So



Solve that and get PG=2.55
CG = CP + PG
CG = 1.2+2.55
CG = 3.75

Triangles GTP and GCE are similar right triangles for they have
a common acute angle at G.  And since right triangles OBP and GTP 
are similar, GCE and OBP are similar, so




Solve that and get EC=2

Edwin