AB and AC are tangents to a circle with centre O. A is the external point of the circle. The line AO intersect the chord BC at D. The measure of the $$\angle$$BDO is:

As shown in the figure,

In $$\triangle\ $$ABO and $$\triangle\ $$ACO

AB = AC, OB = OC and AO is a common side

$$=$$>  $$\triangle\ $$ABO and $$\triangle\ $$ACO are congruent traingles

$$=$$>  $$\angle$$BAO = $$\angle$$CAO

$$=$$>  AO bisects the chord BC

Hence AO is the perpendicular bisector of BC

$$=$$>  $$\angle$$ADB = $$\angle$$ADC = 90$$^\circ$$

From the figure

$$\angle$$ADB + $$\angle$$BDO = 180$$^\circ$$

$$=$$>  90$$^\circ$$ + $$\angle$$BDO = 180$$^\circ$$

$$=$$>  $$\angle$$BDO = 90$$^\circ$$

Hence, the correct answer is Option C