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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Proof of Circle Theorems

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