Question : AB and AC are tangents to a circle with centre O. A is the external point of the circle. The line AO intersects the chord BC at D. The measure of the $\angle$BDO is:
Option 1: 60°
Option 2: 90°
Option 3: 45°
Option 4: 75°
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Correct Answer: 90°
Solution : AO intersects the chord BC at D. In $\triangle$ ABD and $\triangle$ ACD AB = AC = tangents from an exterior point A $\angle$ BAD = $\angle$CAD AD is common. So, $\angle$ BAD $\cong$ $\angle$CAD $\angle$ BDA and $\angle$ ADC are linear pair and equal. $\angle$ BDA + $\angle$ ADC = 180° ⇒$\angle$ BDA = 90° $\angle$ BDA + $\angle$BDO = 180° Therefore, $\angle$BDO = 180° – 90° = 90° Hence, the correct answer is 90°.
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