If DF⊥AC is in F, then in △ABD and △ADF, there are: AD=AD, ∠BAD=∠DAF, ∠DBA=∠DFA, ∴△ABD≌△ADF.
∴ There is BD=DF, and ∵BD is ⊙D radius, so DF is also ⊙D radius, that is, point F is on AC.
∴AC is the tangent of⊙ D.
(2) According to (1), AC is the tangent of ⊙D, with AB=AF.
In △DFC and △EBD, there are ∠ Abd = ∠ DFC = 90, BD=DF, and DE=DC.
∴△DFC≌△EBD has FC=BE.
∫AC = AF+FC
∴AB+EB-AC=AB+EB-AF-FC=0
That is AB+EB=AC.
This question is mainly about congruent triangles's knowledge.