Circles: Angle and Cyclic Properties — chapter notes

• A chord joins two points of a circle; a chord through the centre is a diameter.
• A chord cuts the circle into a minor arc and a major arc; the two regions it cuts off are segments.
• An arc subtends an angle at a point when its two ends are joined to that point.
• Angles are measured at the centre (the central angle) or at a point of the circumference (an inscribed angle).

Example: a chord $PQ$ with $\angle POQ = 80^\circ$ at the centre; the major arc is on the far side of the centre.

• Theorem: the angle subtended by an arc at the centre is double the angle it subtends at any point on the remaining part of the circle: $\angle AOB = 2\angle ACB$. The proof joins $PO$ to $Q$ and uses the exterior angles of two isosceles triangles.
• For a point on the minor arc, use the reflex angle at the centre: $\angle AQB = \dfrac{1}{2}(360^\circ - \angle AOB)$. Points on opposite sides of a chord see it under supplementary angles.
• Then use $OA = OB$ to find the base angles of $\triangle OAB$.

Example: $\angle ACB = 38^\circ$ gives $\angle AOB = 76^\circ$ and $\angle OAB = \dfrac{180^\circ - 76^\circ}{2} = 52^\circ$.

• Equal chords of a circle subtend equal angles at the centre; equal arcs do the same.
• Conversely, equal angles at the centre give equal chords and equal arcs.
• Equal chords (and arcs) also subtend equal angles at the circumference.

Example: $AB = BC$ and $\angle AOB = 64^\circ$: $\angle BOC = 64^\circ$, $\angle AOC = 128^\circ$, $\angle ADC = 64^\circ$ for $D$ on the major arc and $\angle ABC = \dfrac{360^\circ - 128^\circ}{2} = 116^\circ$.

• Angles subtended by the same chord at points of the same side of it (the same segment) are equal: $\angle APB = \angle AQB$.
• Each equals half the angle at the centre on the other side of the chord.
• Points on the opposite side of the chord give the supplementary angle.
• Not in the ICSE 2027 syllabus: the proofs that angles in the same segment are equal and of the exterior angle property of a cyclic quadrilateral (section 7). Both results are used.

Example: chords $AC$ and $BD$ cut at $E$, $\angle ABD = 58^\circ$ and $\angle DAC = 35^\circ$: $\angle ACD = 58^\circ$, $\angle DBC = 35^\circ$ and $\angle ADC = 180^\circ - 35^\circ - 58^\circ = 87^\circ$.

• The angle in a semicircle is a right angle: if $AB$ is a diameter and $P$ is on the circle, $\angle APB = 90^\circ$.
• Proof: $OA = OP = OB$, so $\angle OAP = \angle OPA = x$ and $\angle OBP = \angle OPB = y$; the angles of $\triangle ABP$ give $2x + 2y = 180^\circ$, so $\angle APB = x + y = 90^\circ$.
• It is the centre theorem with a central angle of $180^\circ$. A chord that subtends $90^\circ$ at the circumference is a diameter.

Example: $\angle CAB = 35^\circ$ in a semicircle gives $\angle ABC = 180^\circ - 90^\circ - 35^\circ = 55^\circ$.

• A quadrilateral is cyclic when its four vertices lie on a circle; the vertices are concyclic.
• Theorem: the opposite angles of a cyclic quadrilateral are supplementary: $\angle A + \angle C = 180^\circ$ and $\angle B + \angle D = 180^\circ$. The proof joins $B$ and $D$ to the centre: $\angle A = \dfrac{x}{2}$ and $\angle C = \dfrac{360^\circ - x}{2}$.
• A cyclic parallelogram is a rectangle; a rectangle and a square are always cyclic; a rhombus is cyclic only if it is a square.
• Algebra: set the sum of opposite angles equal to $180$ and solve.

Example: $\angle A = (3x + 5)^\circ$ and $\angle C = (2x - 15)^\circ$ give $5x - 10 = 180$, $x = 38$.

• The exterior angle of a cyclic quadrilateral (a side produced) equals the interior opposite angle: $\angle DCE = \angle BAD$.
• Use the exterior angle property to find angles without first finding the interior angle.
• Converse: if a pair of opposite angles is supplementary, or an exterior angle equals the interior opposite angle, the four points are concyclic; equal angles on the same side of a chord also give concyclic points.
• When sides $AB$ and $DC$ are produced to meet at $E$, the triangle $EBC$ gives the angle at $E$.

Example: $\angle ADC = 70^\circ$ and $\angle DAB = 75^\circ$: $\angle EBC = 70^\circ$, $\angle ECB = 75^\circ$ and $\angle BEC = 180^\circ - 70^\circ - 75^\circ = 35^\circ$.