In the diagram, O is the centre of the circle. RSTV is a cyclic quadrilateral. Use the diagram to prove the theorem which states that the opposite angles of a cyclic quadrilateral are supplementary, i.e. \(\hat{R} + \hat{T} = 180^\circ\)
About this question
Question 9.1 of Mathematics Grade 12 Paper 2, NSC Examination (May/June) (DBE, May/June 2026). It tests the Term 3 ATP topic Euclidean Geometry.
- TermTerm 3
- ATP topicEuclidean Geometry
- PaperPaper 2
- Marks5 marks
9.1 O is the centre of the circle. RSTV is a cyclic quadrilateral, with R and T opposite each other.
9.2 A, B, C, D and E lie on a circle. BA is produced to F. \(\hat{B}_2 = 27^\circ\) and \(\hat{C} = 88^\circ\). AE, ED, DC, BC, BE and BD are drawn.
Angle labels in 9.2: at B, \(\hat{B}_1 = \angle ABE\), \(\hat{B}_2 = \angle EBD\), \(\hat{B}_3 = \angle DBC\). At E, \(\hat{E}_1 = \angle AEB\) and \(\hat{E}_2 = \angle BED\). At D, \(\hat{D}_1 = \angle EDB\) and \(\hat{D}_2 = \angle BDC\). At A, \(\hat{A}_1 = \angle BAE\) and \(\hat{A}_2 = \angle FAE\) (outside the circle). \(\hat{C} = \angle BCD = 88^\circ\).
Construction: draw radii OS and OV. Let \(\hat{O}_1\) be the angle at O facing T and \(\hat{O}_2\) the reflex angle at O facing R.
- \(\hat{O}_2 = 2\hat{R}\) (\(\angle\) at centre = 2 \(\times\) \(\angle\) at circumference)
- \(\hat{O}_1 = 2\hat{T}\) (\(\angle\) at centre = 2 \(\times\) \(\angle\) at circumference)
- \(\hat{O}_1 + \hat{O}_2 = 360^\circ\) (\(\angle\)s round a pt)
- \(2\hat{R} + 2\hat{T} = 360^\circ\)
- \(\therefore \hat{R} + \hat{T} = 180^\circ\)
\(\hat{R} + \hat{T} = 180^\circ\)
More Questions On Euclidean Geometry
\(\dfrac{QS}{QR} = \dfrac{QZ}{QS}\) (\(\triangle QSZ \,|||\, \triangle QRS\))
\(QS^2 = QZ \cdot QR\)
But \(PQ^2 = RQ \cdot QZ\) (proved in 11.2.2)
\(\therefore PQ^2 = QS^2\)
\(\therefore PQ = QS\)
- \(\hat{S}_1 + \hat{S}_2 = \hat{D} = 90^\circ - x\) (\(\angle\)s in the same segment)
- In \(\triangle BKS\): \(\hat{K}_3 = 90^\circ\) (given \(AS \perp BD\))
- \(\hat{B}_2 = 180^\circ - 90^\circ - (90^\circ - x)\) (sum of \(\angle\)s in \(\triangle\))
\(\hat{B}_2 = x\)
In \(\triangle DFE\) and \(\triangle TFO\):
- \(\hat{D}_3 = \hat{T}_3\) (ext \(\angle\) of cyclic quad)
- \(\hat{F}_4 = \hat{C}_2\) (tan chord theorem)
- \(\hat{C}_2 = \hat{F}_2\) (\(\angle\)s opp equal sides, \(OC = OF\))
- So \(\hat{F}_4 = \hat{F}_2\)
- \(\hat{E} = \hat{O}_2\) (3rd \(\angle\) of \(\triangle\))
- \(\therefore \triangle TFO \,|||\, \triangle DFE\) (\(\angle\angle\angle\))
In \(\triangle AGF\) and \(\triangle ABC\):
\(A\hat{G}F = \hat{B}_2\) (\(\angle\)s in the same seg)
\(A\hat{F}G = \hat{C}_1 = x\) (proved in 11.1)
\(\hat{A}_3 = \hat{A}_2\) (sum of \(\angle\)s in \(\triangle\))
\(\therefore \triangle AGF \,|||\, \triangle ABC\) (\(\angle\angle\angle\))
\(\hat{P} = \hat{R}_2 = 68^\circ\) (ext \(\angle\) of cyclic quad)
Construction: draw KO and produce it to a point on the arc TP. Label the angles \(\hat{O}_1\) (next to T), \(\hat{O}_2\) (next to P), \(\hat{K}_1 = T\hat{K}O\) and \(\hat{K}_2 = O\hat{K}P\).
\(\hat{O}_1 = \hat{K}_1 + \hat{T}\) (ext \(\angle\) of \(\triangle\))
\(\hat{K}_1 = \hat{T}\) (\(\angle\)s opp equal sides; \(OK = OT\), radii)
\(\therefore \hat{O}_1 = 2\hat{K}_1\)
Similarly \(\hat{O}_2 = \hat{K}_2 + \hat{P}\) (ext \(\angle\) of \(\triangle\)) and \(\hat{K}_2 = \hat{P}\) (\(\angle\)s opp equal sides), so \(\hat{O}_2 = 2\hat{K}_2\)
\(\hat{O}_1 + \hat{O}_2 = 2\hat{K}_1 + 2\hat{K}_2 = 2(\hat{K}_1 + \hat{K}_2)\)
\(\therefore T\hat{O}P = 2\,T\hat{K}P\)