Circle theorems, with proofs
Tonight's idea
Circle theorems are one theorem and a few short consequences.
Take a circle with centre \(O\) and two points \(A\), \(B\) on it. The angle \(\angle AOB\) at the centre is the central angle. Pick any third point \(C\) on the circle, on the far side from the arc \(AB\): \(\angle ACB\) is an inscribed angle on the same arc.
Inscribed angle theorem. \(\angle ACB = \tfrac12 \angle AOB\). The angle at the centre is twice the angle at the circumference, when both stand on the same arc.
Proof. Draw the diameter \(CD\), splitting \(\theta\) into \(\alpha\) and \(\beta\) as in the figure. \(OA = OC\) (radii), so triangle \(OAC\) is isosceles: \(\angle OCA = \angle OAC = \alpha\). The exterior angle \(\angle AOD\) of that triangle equals the sum of the two remote interior angles: \(2\alpha\). In the same way \(\angle BOD = 2\beta\). Adding, \(\angle AOB = 2(\alpha + \beta) = 2\theta\). (If \(O\) lies outside the angle, subtract instead of adding.) \(\square\)
Consequences, each a one-liner now:
- Angles in the same segment are equal. Both are half the same central angle.
- Thales: the angle in a semicircle is 90°. A diameter gives a central angle of 180°.
- Opposite angles of a cyclic quadrilateral add to 180°. Their two arcs make the whole circle, so the central angles add to 360°; halve.
- A tangent is perpendicular to the radius at the point of contact, and the perpendicular from the centre bisects a chord (two congruent right triangles).
Watch one: the figure above, with \(\angle AOB = 100°\)
Then \(\angle ACB = 50°\) for \(C\) anywhere on the major arc (the longer way round from \(A\) to \(B\)): it does not matter where \(C\) is, which is the "same segment" fact. Put a point \(D\) on the minor arc (the shorter way). \(ACBD\) is a cyclic quadrilateral, so \(\angle ADB = 180° - 50° = 130°\). Alternatively, \(D\) sees arc \(AB\) from the other side, where the central angle is the reflex angle \(360° - 100° = 260°\), and half of that is 130°. Both routes agree.
Your turn
- \(O\) is the centre and \(\angle AOC = 140°\). \(B\) is on the major arc \(AC\) and \(D\) is on the minor arc. Find \(\angle ABC\) and \(\angle ADC\), and name the theorem you used for each.
- Triangle \(PQR\) is inscribed in a circle with \(PR\) a diameter, and \(\angle QPR = 37°\). Find the other two angles of the triangle. Which theorem makes this a one-step problem?
- \(ABCD\) is a cyclic quadrilateral with \(\angle A = 2x + 10\) and \(\angle C = 3x - 30\) (in degrees). Find \(x\) and both angles. If \(\angle B = 115°\), find \(\angle D\).
- (a) From a point \(P\) outside a circle with centre \(O\), the tangent \(PT\) has length 12 and \(PO = 13\). Find the radius. (b) A chord of length 24 is drawn in a circle of radius 13. How far is the chord from the centre? Draw the right triangle you used in each part.
Stretch: a five-pointed star
Draw a regular pentagon inscribed in a circle and join every vertex to every other: a pentagram. Each sharp point of the star is an inscribed angle. Which arc does it stand on, and how big is that arc? Find the angle at each point, and the total of all five points.
Answers
- \(\angle ABC = 70°\) (inscribed angle is half the central angle). \(\angle ADC = 110°\): either opposite angles of the cyclic quadrilateral \(ABCD\) add to 180°, or \(D\) sees the reflex central angle \(220°\), half of which is 110°.
- \(\angle PQR = 90°\) by Thales (angle in a semicircle), so \(\angle PRQ = 180° - 90° - 37° = 53°\).
- \(2x + 10 + 3x - 30 = 180\), so \(5x = 200\) and \(x = 40\). \(\angle A = 90°\) and \(\angle C = 90°\) (so \(BD\) is a diameter, by the converse of Thales). \(\angle D = 180° - 115° = 65°\).
- (a) \(OT \perp PT\), so triangle \(OTP\) is right-angled at \(T\): \(r^2 = 13^2 - 12^2 = 25\), \(r = 5\). (b) The perpendicular from the centre bisects the chord into two halves of 12; with hypotenuse 13 the distance is \(\sqrt{169 - 144} = 5\). Same triangle, read the other way round.
Stretch. The five vertices cut the circle into five equal arcs of \(360°/5 = 72°\). A point of the star is an inscribed angle whose two sides go to the two vertices on the opposite side, and the arc between those two vertices (not containing the point's vertex) is one of the five arcs: 72°. So each point is \(36°\), and the five points total \(180°\). (A seven-pointed star that joins each vertex to the two farthest from it also totals 180°: each point stands on one arc of \(360°/7\).)