The answer is as follows: (1) Because DE is the tangent of a circle, the angle ACD= the angle AEC (the chord tangent angle is equal to the circumferential angle of the arc it encloses). Because AE is the diameter of a circle, angle ACE=90 degrees and angle D=90 degrees, angle EAC= angle CAD (complementary angles of equal angles are equal), so BC = CF

(2)AD=6, DE=8, so AE= 10, and the intersection point C is CM perpendicular to AE and M, so CM=CD (the distance from the point on the bisector of the angle is equal to both sides of the angle), and it is easy to get EC=5 and CD=3 by using Pythagorean theorem. In triangle BEC and triangle CEA, angle E= angle E, and angle ECB= So EC2=BE*AE so BE=2.5.

(3) Auxiliary line for the second question: When the intersection point C is CM perpendicular to AE and point M, it is easy to prove that the triangle MCB is all equal to the triangle DCF, so DF=EM, AM=AD, so AB=AM+MB=AD+EM=AD+DF=AF+2DF.

I hope my answer is helpful to you. I mainly apply the basic properties of congruence and similarity to investigate the tangent angle theorem in a circle and the property that the circumferential angle opposite to the diameter is a right angle.