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.
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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 !!
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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