# Proof of Circle Theorems

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

##### Circle TheoremsHelp VideoMore on CirclesMore on Angles

Drag the statements proving the theorem into the correct order.

 Similarly ∠AOC = 180° – 2 x ∠OCAOB = 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 DTo prove: ∠ABD = ∠ACD Similarly in triangle BCO ∠OCB = ∠OBCTo 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 ∠CDATo 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 ∠CBDTo prove ∠CAB = ∠CBD∠OBC = ∠OCB (equal angles in isosceles triangle OBC)Construct the radii OB and OC2 x ∠OBC + ∠COB = 2(∠OBC + ∠CBD) (from [2] and [3] above)
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