Question 1078647
<pre><font size=4><b>
{{{drawing(400,4200/13,-.5,25.5,-1,20,
line(9,0,9,9), line(21,0,21,4), line(21,4,9,9),
locate(9,10,A),locate(9,0,B),locate(15,0,F),locate(21,5,D),locate(225/13-.5,1.3+72/13,E), locate(21,0,C),locate(21.2,2.7,4),locate(8.2,5,9),
line(15,0,20,12),line(-2,0,27,0),circle(9,9,9),circle(21,4,4) )}}}

Now we draw in line DG (in green) parallel and equal to BC.
It intersects EF at I.  DG divides radius AB = 9 into AG and BG. 
Since CD = BG = 4, AG = AB-BG = 9-4 = 5 


{{{drawing(400,4200/13,-.5,25.5,-1,20,
line(9,0,9,9), line(21,0,21,4), line(21,4,9,9),locate(8.2,2.5,4),locate(8.2,7,5),
locate(9,10,A),locate(9,0,B),locate(15,0,F),locate(21,5,D),locate(225/13-.5,1.3+72/13,E), locate(21,0,C),locate(21.2,2.7,4),locate(8.2,4.8,G),
line(15,0,20,12),  locate(16,5,I),



green(line(9,4,21,4)),
line(-2,0,27,0),circle(9,9,9),circle(21,4,4) )}}}



Now we use the Pythagorean theorem on right triangle AGD:
Hypotenuse = AD = radius AE + radius DE = 9+4 = 13

DG<sup>2</sup> + AG<sup>2</sup> = AD<sup>2</sup>
DG<sup>2</sup> + 5<sup>2</sup> = 13<sup>2</sup>
DG<sup>2</sup> + 25 = 169
DG<sup>2</sup> = 144
DG = 12

Now we have BC = DG = 12.

Next, we draw in EH parallel to AB and CD and perpendicular
to BC and G.  DG and EH are perpendicular and intersect at J 

{{{drawing(400,4200/13,-.5,25.5,-1,20, locate(16,5,I),
line(9,0,9,9), line(21,0,21,4), locate(17.5,4,J), line(21,4,9,9),locate(8.2,2.5,4),locate(8.2,7,5),
locate(9,10,A),locate(9,0,B),locate(15,0,F),locate(21,5,D),locate(225/13-.5,1.3+72/13,E), locate(21,0,C),locate(21.2,2.7,4),locate(8.2,4.8,G),
line(15,0,20,12),

green(line(225/13,0,225/13,72/13)),locate(225/13,0,H),
green(line(9,4,21,4)),
line(-2,0,27,0),circle(9,9,9),circle(21,4,4) )}}}

By similar triangles EJD and AGD,

{{{EJ/(AG)="DE"/"AD"}}}

{{{"EJ"/5=4/13}}}

{{{13*"EJ"=20}}}

{{{EJ=20/13}}}

HJ = CD = 4

Therefore EH = EJ+HJ = {{{20/13+4=20/13+52/13=72/13}}}

Triangles IJE and EJD are similar since they are right
triangles with a common angle at D.

Triangles IJE and FHE are similar

Therefore triangles FHE and AGD are similar

{{{"EF"/"EH"="AD"/"DG"}}}

{{{"EF"/(72/13)=13/12}}}

{{{EF*expr(13/72)=13/12}}}

Divide both sides by 13

{{{EF/72=1/12}}}

{{{EF=72/12}}}

{{{EF=6}}}

Edwin</pre>