If the given figure, $$\angle$$ACB + $$\angle$$BAC = 80$$^\circ$$; $$\angle$$BDE = 35$$^\circ$$; $$\angle$$BCE = 45$$^\circ$$, then the marked angle $$\angle$$CED is:

Given,

$$\angle$$ACB + $$\angle$$BAC = 80$$^\circ$$

$$\angle$$BDE = 35$$^\circ$$

$$\angle$$BCE = 45$$^\circ$$

In $$\triangle$$ABC,

$$\angle$$ACB + $$\angle$$BAC + $$\angle$$ABC = 180$$^\circ$$

$$=$$>  80$$^\circ$$ + $$\angle$$ABC = 180$$^\circ$$

$$=$$>  $$\angle$$ABC = 100$$^\circ$$

From the figure,

$$\angle$$ABC + $$\angle$$CBD = 180$$^\circ$$

$$=$$>  100$$^\circ$$ + $$\angle$$CBD = 180$$^\circ$$

$$=$$>  $$\angle$$CBD = 80$$^\circ$$

In $$\triangle$$FBD,

$$=$$>  $$\angle$$FBD + $$\angle$$BDF+ $$\angle$$BFD = 180$$^\circ$$

$$=$$>  $$\angle$$CBD + $$\angle$$BDE + $$\angle$$BFD = 180$$^\circ$$

$$=$$>  80$$^\circ$$ + 35$$^\circ$$ + $$\angle$$BFD = 180$$^\circ$$

$$=$$>  115$$^\circ$$ + $$\angle$$BFD = 180$$^\circ$$

$$=$$>  $$\angle$$BFD = 65$$^\circ$$

In $$\triangle$$CFE,

$$\angle$$BFD is the exterior angle at point F

$$=$$>  $$\angle$$BFD = $$\angle$$FCE + $$\angle$$CEF

$$=$$>  $$\angle$$BFD = $$\angle$$BCE + $$\angle$$CEF

$$=$$>  65$$^\circ$$ =  45$$^\circ$$ + $$\angle$$CEF

$$=$$>  $$\angle$$CEF = 20$$^\circ$$

From the figure,

$$\angle$$CEF + $$\angle$$CED = 180$$^\circ$$

$$=$$>  20$$^\circ$$ + $$\angle$$CED = 180$$^\circ$$

$$=$$>  $$\angle$$CED = 160$$^\circ$$

Hence, the correct answer is Option B