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Proof of Circle Theorems

Arrange the stages of the proofs for the standard circle theorems in the correct order.

Circle Theorems Help Video More on Circles More on Angles

Drag the statements proving the theorem into the correct order.

	Similarly ∠AOC = 180° – 2 x ∠OCA OB = OC (radii of circle) ∠BOA = 2∠BCA Q.E.D. Construct radius OC ∠COB = 180° – 2 x ∠BCO (Angle sum of triangle OBC) To prove: ∠BOA = 2∠BCA ∠BCO = ∠OBC (equal angles in isosceles triangle) ∠BOA = 2(∠BCO + ∠OCA) ∠BOA = 360° – (180° – 2 x ∠BCO + 180° – 2 x ∠OCA) ∴ OBC is an isosceles triangle
	∴ 2 x ∠ABD = 2 x ∠ACD ∠AOD = 2 x ∠ABD (angle at centre twice angle at circumference) ∠AOD = 2 x ∠ACD (angle at centre twice angle at circumference) ∠ABD = ∠ACD Q.E.D. Construct radii from A and D To prove: ∠ABD = ∠ACD
	Similarly in triangle BCO ∠OCB = ∠OBC To prove: ∠ABC = 90° ∴ ∠OAB = ∠OBA (equal angles in isosceles triangle ABO) ∠OAB + ∠OBA + ∠OCB + ∠OBC = 180° (Angle sum of triangle ABC) OA = OB (radii) ∴ 2(∠OBA + ∠OBC) = 180° ∴ ABO is an isosceles triangle (two equal sides) ∠ABC = 90° Q.E.D. Construct the radius OB ∴ ∠OBA + ∠OBC = 90°
	The obtuse and reflex angles at O add up to 360° (angles at a point) Similarly the obtuse angle AOC = 2 x ∠CDA To prove ∠ABC + ∠CDA = 180° ∴ 2 x ∠ABC + 2 x ∠CDA = 360° Reflex ∠AOC = 2 x ∠ABC (angle at centre twice angle at circumference) ∠ABC + ∠CDA = 180° Q.E.D. Construct the radii OA and OC
	2 x ∠CAB = 2 x ∠CBD (from [1] above) ∠OBC + ∠CBD = 90° (angle between radius and tangent) [2] ∠CAB = ∠CBD Q.E.D. 2 x ∠OBC + ∠COB = 180° (angle sum of triangle) [3] Obtuse ∠COB = 2 x ∠CAB (angle at centre twice angle at circumference) [1] ∠COB = 2 x ∠CBD To prove ∠CAB = ∠CBD ∠OBC = ∠OCB (equal angles in isosceles triangle OBC) Construct the radii OB and OC 2 x ∠OBC + ∠COB = 2(∠OBC + ∠CBD) (from [2] and [3] above)

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Proof of Circle Theorems

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Curriculum Reference

See the National Curriculum page for links to related online activities and resources.

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