Question

please help me to solve the question​

Answer 1

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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Answer 2

$\huge \fbox {ANSWER}$

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