In $$\triangle$$ABC, D is a point on BC such that $$\angle$$BAD = $$\frac{1}{2} \angle$$ADC and $$\angle$$BAC = 77$$^\circ$$ and $$\angle$$C = 45$$^\circ$$. What is the measure of $$\angle$$ADB?

$$\angle$$ADC = x

Given, $$\angle$$BAD = $$\frac{1}{2} \angle$$ADC

$$\angle$$BAD = $$\frac{\text{x}}{2}$$

$$\angle$$DAC = 77 - $$\frac{\text{x}}{2}$$

From triangle ADC,

$$\angle$$ADC + $$\angle$$ACD + $$\angle$$DAC = 180$$^\circ$$

x + 45$$^\circ$$ + 77 - $$\frac{\text{x}}{2}$$ = 180$$^\circ$$

$$\frac{\text{x}}{2}$$ = 58$$^\circ$$

x = 116$$^\circ$$

$$\angle$$ADB = 180$$^\circ$$ - x

= 180$$^\circ$$ - 116$$^\circ$$

= 64$$^\circ$$

Hence, the correct answer is Option B