E is the midpoint of the arc CD.

∴2∠ACE=2∠DCE=∠DOE=∠EOC=2∠ECA'

∴A and A coincide.

∴AC is the tangent of circle O.

(2)? Let the CD and EO go to H? AD=32/5BD= 18/5CD=24/5

∫e is the midpoint of arc CD, and o is the midpoint of BC.

∴OE||AB

∴oh= 1/2bd=9/5oe=od= 1/2bc=3

∴EH=6/5？ CH= 12/5

If e is used as EF||CD, DF=EH=6/5.

∴AF=32/5-6/5=26/5

∴ AE = (26/5) 2+(12/5) 2 = 2/5 * 205 under the radical sign.