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0580/22 Feb/Mar 2025, Question 14: Cyclic Quadrilaterals

Cambridge IGCSE Mathematics 0580 · Paper 22 (Extended) · Non-calculator · 5 marks

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The question, in short

Using a circle through A, B, C, D, E with tangent FG at C and given angles BAD = 110°, ADB = 20°, BEC = 45°, find angles BCD, DBC and DCG, giving a reason for BCD.

Summarised in our own words. Use the official paper for the exact wording and the diagram.

Part (a) · 2 marks

Find angle BCD and state a geometrical reason for it.

  1. 1Angle BCD = 180 − 110 = 70°
  2. 2because opposite angles of a cyclic quadrilateral add to 180°

Answer70 and opposite angles of a cyclic quadrilateral sum to 180

Part (bi) · 2 marks

Find angle DBC.

  1. 1ABCD is a cyclic quadrilateral, so angle BCD = 180 − 110 = 70° (opposite angles add to 180°)
  2. 2Angle BDC = angle BEC = 45° (angles in the same segment)
  3. 3Angles in triangle BCD add to 180°: angle DBC = 180 − 70 − 45
  4. 4Angle DBC = 65°

Answer65

Part (bii) · 1 mark

Find angle DCG.

  1. 1Angle DCG = angle DBC (alternate segment theorem)
  2. 2Angle DCG = 65°

Answer65

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