∴△ADC∽△ACE,∴CDCE=ACAE，
Similarly, BDBE=ABAE,
AB = AC，
∴CDCE=BDBE,∴BE？ CD=BD？ Ce ... (5 ... (5 points)
(2) Solution: As shown in the figure, from the tangent theorem, FB2=FD? FC，
∫CE∨AB，
∴∠FAD=∠AEC，
∵AB is tangent to B, ∴∠ACD=∠AEC, ∴∠FAD=∠FCA,
△ AFD △ CFA, △ AECF = FDAF, that is, AF2=FD? FC，
∴FB2=AF2, that is, FB=FA, ∴F is the midpoint of line segment AB ... (10)