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Angle at the Centre is Double The Angle at Circumference

Angle at the centre is double The Angle at circumference.

The angle subtended by an arc at the centre is twice the angle subtended at the circumference

The angle at the centre is double the angle at the circumference.

Given : Given a Circle with centre O and a Chord HK

To Prove $large angle HOK=2angle HGK$

Construction: Join OG, OH and OK. Extend HO to J

Proof Let $large angle OGH=y^0$ and $large large angle OGK=x^0$ .

$Delta OGH$ is an isosceles. LengthsOH and OG are both radii of the same circle

$angle OGH=angle OHG = y^0$ [

$Delta OGK$ is an isosceles. Lengths OG and OK are both radii of the same circle.

$angle OGK=angle OKG = x^0$ [

$large angle GOH=180-2y$ [ Angle sum property ] ..............(1)

$large angle GOK=180-2x$ [ Angle sum property ] ..............(2)

$large angle HOK + angle GOK+angle GOH=360^0$ [Angles around a point] ..............(3)

Putting the values from (1) and (2) in (3) we Get

$large angle HOK +180-2x^0+180-2y^0=360^0$

$large angle HOK +360-2x^0 -2y^0=360^0$

$large angle HOK =360^0-360+2x^0 +2y^0$

$large angle HOK =2(x^0 +y^0)$ ..............(4)

$large angle HGK =angle HGO+ angle OGK= x^0 +y^0$ ..............(5)

From (4) and (5 ) we Get

$large angle HOK =2angle HGK$

Hence Proved.

Illustration: If A, B, M are three points on a circle with centre O such that $angle AOM=70^0$ and $angle MOB=140^o,$ Find $angle AMB$

Solution: If A, B, M are three points on a circle with centre O such that $angle AOM=70^0$ and $angle MOB=140^o,$

$angle AOB = angle AOM + angle MOB = 70^0 +140^0 = 210^0$

Reflex of $angle AOB=360^0- 210^0=150^0$

$angle AMB=frac{150^0}{2}=75^0$

As angle at the centre is double the angle and the circumference.

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Angle at the Centre is Double The Angle at Circumference