Proof: (1) Use the "Same Law" certificate. It is easy to prove RT△ABD≌RT△AFD (using the corner theorem, that is, AAS theorem) when the crossing point D is DF⊥AC and the vertical foot is F. So there is DF=BD= radius, so we know that AC is the tangent of circle D.

(2) After passing point D, extend the extension line of FD-AB to the opposite direction of point G to connect BG. Because AC tangent D is at point F and AB tangent D is at point B, there is AB=AF. < BAC = & ltFAG again, so RT△ABC≌RT△AFG (Angle Edge Theorem). So there is

AB+BG=AG=AC，BC=FG。

In RT△EBD and RT△GBD, BD is the same side, DG=FG-DF=BC-BD=DC=DE, so there is RT△EBD≌RT△GBD according to HL theorem. So BE=BG.

So AB+EB=AB+BG=AG=AC.

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