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Question

AB is the diameter of the circle with centre O. OD is parallel to BC and AOD = 60o, Calculate the numerical values of :

(i) ABD ,

(ii) DBC,

(iii) ADC.

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Solution

Join B and D as shown in the diagram.

(i) \angle ABD = \frac{1}{2} \angle AOD = \frac{1}{2} \times 60^o = 30^o (angle subtended by an chord on the center is double that subtended by the same chord on the circumference)c713

(ii) \angle BDA = 90^o (angle in semi circle)

Since \angle OAD = 60^o \Rightarrow \triangle OAD  is equilateral

\angle ODB = 90^o-ODA = 90^o-60^o = 30^o

Since OD \parallel BC

\angle DBC = \angle ODB = 30^o (alternate angles)

(ii) \angle ABC = \angle ABD + \angle DBC = 30^o +30^o = 60^o

\angle ADC = 180^o -\angle ABC = 180^o- 60^o = 120^o ( ABCD is a cyclic quadrilateral)

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