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In the given figure O is the centre of the circle. If diameter AD bisects angle BOC. Prove that angle BOD and angle OBC are complementary angles.

no other details given?


First of all, there was no need of drawing triangle ABC,

Let the point where AD intersects BC be E.
Now in triangles BOE and COE
<BOE = <COE (given that AD bisects <BOC)
OB = OC (radius)
OE is common
Thus triangle BOE is congruent to triangle COE (S.A.S)

Thus <BEO = <CEO = 90°
Now in triangle BEO 
<BEO + <BOE + <OBE = 180⁰
90⁰ + <BOE + <OBE = 180⁰
<BOE + <OBE = 180°- 90° = 90°
or, <BOD + <OBC = 90°

Hope that helps !!

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