Let BE⊥AC be at point E and AD be at point F.

∫∠BAC = 45

∴AE=BE

∠∠FAE+∠C =∠CBE+∠C = 90

∴∠EAF=∠CBE

∴△AEF≌△BCE

∴AF=BC

Let CD = X.

Then DF=4-x

∫tan∠DBF = tan∠CAD

∴(4-x)/4=x/8

The solution is x=8/3.

∴tanC=AD/CD=8/(8/3)=3