(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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