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Solution :
O is the circumcentre of the circle ( And should also be the centroid of ∆ ABC , but isn't mentioned ) .
∆ ABC is an isosceles ∆ , AB = AC.
Angle ABC = Angle ACB = 67.5 °
Therefore ;
Angle BOC lying at the centre of the circle is double the angle subtended on the arc ;
Angle BOC = 2 × Angle BAC
Angle BAC -
> [ 180 - 2 × 67.5 ]
> 180 - 135
> 45°
Therefore angle BOC = 2 × 45 = 90°
∆ OBC is a right angled triangle.
OB = OC = Radius of circle .
∆ OBC is an isosceles right angled triangle .
Angle OBC = Angle OCB = ( 180 - 90 )/2 = 45°
Therefore ;
Angle OBC = 45°
Angle BAC = 45° [ Proved before ]
Adding them ;
Angle OBC + Angle BAC = 90°
Hence Proved !
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Verified answer
[tex]\huge \fbox {ANSWER}[/tex]
Here, O is the circumcentre of circle.
Angle ABC = Angle BAC = 67.5⁰
Angle BOC = Angle BAC × 2
BAC = 180 - 2 × 67.5
BAC = 45⁰
BOC = 2(45) = 90⁰
∆OBC is a right angled triangle
OB and OC are radii of circle
Therefore,
OB = OC
Angle OBC = 180 - 90/2 = 45⁰
Angle BAC = 45⁰
OBC + BAC = 90
45 + 45 = 90
90 + 90
Hence,
Proved