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 Now we use the Pythagorean theorem on right triangle AGD: Hypotenuse = AD = radius AE + radius DE = 9+4 = 13 DG2 + AG2 = AD2 DG2 + 52 = 132 DG2 + 25 = 169 DG2 = 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 By similar triangles EJD and AGD, HJ = CD = 4 Therefore EH = EJ+HJ = 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 Divide both sides by 13 Edwin