In the given figure, AD is bisector of angle $$\angle$$CAB and BD is bisector of angle $$\angle$$CBF. If the angle at C is 34$$^\circ$$, the angle $$\angle$$ADB is:

Given,

$$\angle$$C = 34$$^\circ$$

AD is bisector of angle $$\angle$$CAB

$$=$$> $$\angle$$ACE = $$\angle$$BAE

Let $$\angle$$ACE = $$\angle$$BAE = x

BD is bisector of angle $$\angle$$CBF

$$=$$> $$\angle$$CBD = $$\angle$$FBD

Let $$\angle$$CBD = $$\angle$$FBD = y

In $$\triangle\ $$ABC, $$\angle$$CBF is the exterior angle at B

$$=$$>  2y = 34$$^\circ$$ + 2x

$$=$$>  y = 17$$^\circ$$ + x

In $$\triangle\ $$ABC,

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

$$=$$>  $$\angle$$ABC + 2x + 34$$^\circ$$ = 180$$^\circ$$

$$=$$>  $$\angle$$ABC = 146$$^\circ$$- 2x

In $$\triangle\ $$ABD,

$$=$$>  $$\angle$$ABD + $$\angle$$BAD + $$\angle$$ADB = 180$$^\circ$$

$$=$$>  146$$^\circ$$- 2x + y + x + $$\angle$$ADB = 180$$^\circ$$

$$=$$>  146$$^\circ$$- x + 17$$^\circ$$ + x + $$\angle$$ADB = 180$$^\circ$$

$$=$$>  163$$^\circ$$ + $$\angle$$ADB = 180$$^\circ$$

$$=$$>  $$\angle$$ADB = 17$$^\circ$$

Hence, the correct answer is Option B