Warm-up
Three quick ones from earlier work. Try each in your head or on paper before you open the answer.
- From the previous lesson: the angle at the centre of a circle on an arc is 84°. Find the angle at the circumference on the same arc, on the major arc.
Answer
42°, because it is half. - From the previous lesson: a cyclic quadrilateral has an angle of 71°. Find the opposite angle.
Answer
109°, because opposite angles add to 180°. - From Year 9: a triangle has angles of 38° and 67°. Find the third angle.
Answer
75°, because the three add to 180°.
Words used in this lesson
- converse
- the statement turned round: if B then A instead of if A then B. For example: if it is a square, it has four sides; the converse is false.
- concyclic
- lying on one circle. For example: four posts round a circular pond.
- contradiction
- a result that cannot be true at the same time as what you assumed. For example: assume it is raining and dry at once.
- subtend
- to make an angle at a point from the ends of a line. For example: the harbour mouth subtends an angle at the surveyor.
- diameter
- a chord through the centre; the longest chord. For example: the width of a round table.
Where can the surveyor stand?
Two posts A and B mark the ends of a harbour mouth. A surveyor at P measures the angle APB and finds 50°. She wants to stand somewhere else and see the harbour mouth at exactly 50° again, on the same side of AB.
Predict first: what shape do all the places she could stand make? Sketch your guess and write a reason, then use the first figure.
P is on the circle, so angle APB = 50.0°, the same as at every other point of the major arc.
Show me the steps
- P at the top of the circle, on it: angle APB = 50.0°. Slide P round the major arc and the angle stays 50.0°: that is the same-segment theorem.
- Move P 10% inside the circle at the top: the angle grows to 52.8°. Inside, P is closer to the chord's ends and sees the chord at a wider angle.
- Move P 20% inside at the right-hand end: 64.7°. The further inside, the larger the angle.
- Move P 20% outside the circle at the top: the angle is 45.1°. Outside, P is further away, and sees the chord at a narrower angle. Exactly 50° happens only on the circle.
The theorems run both ways. (1) If angle APB = 90°, then P lies on the circle with diameter AB. (2) If P and Q are on the same side of AB and angle APB = angle AQB, then A, B, P and Q lie on one circle. (3) If opposite angles of a quadrilateral add to 180°, it is cyclic.
Proof of (2), by contradiction. Draw the circle through A, B and P. Suppose Q is not on it. Then Q is inside or outside the circle. Inside, the angle AQB is greater than APB; outside, it is smaller. Either way angle AQB ≠ angle APB, which contradicts the given equality, so Q is on the circle.
a = 8, b = 6: hypotenuse 10.0 cm, and the midpoint M of the hypotenuse is 5.0 cm from A, from B and from the right-angled corner C.
Show me the steps
- a = 8, b = 6: the hypotenuse is 10.0 and its midpoint M is 5.0 from A, from B, and from the right-angled corner C. A circle with centre M and radius 5.0 passes through all three.
- Change a to 12 with b = 6: the hypotenuse is 13.4, M is 6.7 from each corner, and the circle is bigger. The distance from M to C still equals the other two.
- Change b to 12 with a = 6: the hypotenuse is 13.4 again, a mirror image, and again MC = MA = MB = 6.7.
- For every a and b the right angle's vertex lies on the circle on the hypotenuse as diameter. The converse is the useful form: a 90° angle subtended by AB puts C on that circle.
Equal angles are not always enough. Equal angles do not always put four points on one circle. A point P above AB and a point Q below AB can both see AB at 50° without being concyclic. Check where the points sit before you use the converse.
Worked examples
1. Prove that if angle ACB = 90° then C lies on the circle with diameter AB.
- Let M be the midpoint of AB. Extend CM to D so that MD = CM. The diagonals AB and CD of the quadrilateral ACBD bisect each other, so ACBD is a parallelogram.
- Angle ACB = 90°, so the parallelogram has a right angle, and is a rectangle.
- The diagonals of a rectangle are equal, so CD = AB and MC = ½CD = ½AB = MA = MB.
- So C is the same distance from M as A and B are: C is on the circle with centre M and diameter AB.
2. Points A, B, C, D are such that angle ABD = 38° and angle ACD = 38°, with B and C on the same side of AD and the points in the order A, B, C, D round the circle. Angle BAC = 27°. Find angle BDC.
- B and C see AD at the same angle from the same side, so A, B, C, D are on one circle.
- In the order A, B, C, D, the points A and D are both on the arc not between B and C, so angles BAC and BDC stand on the same chord BC from the same side. So angle BDC = angle BAC = 27°.
3. Show that if opposite angles of a convex quadrilateral ABCD add to 180° then A, B, C, D are concyclic.
- Draw the circle through A, B and D. Take a point C′ on its arc BD that is on the opposite side of BD from A. Then ABC′D is cyclic, so angle BC′D = 180° − angle A.
- Angle BCD = 180° − angle A as well, because angle A + angle C = 180°. So angle BCD = angle BC′D.
- C and C′ are on the same side of BD, the side opposite A, because ABCD is convex. By the converse of the same-segment theorem, B, D, C and C′ are concyclic. The circle through B, D and C′ is the one through A, so C is on it.
Your go. The first step is given. In a quadrilateral PQRS, angle P = (3x + 10)° and angle R = (2x + 20)°. Given that PQRS is cyclic, find angle P.
- P and R are opposite angles of a cyclic quadrilateral, so they add to 180°.
- Write (3x + 10) + (2x + 20) = 180 and solve for x, then find angle P.
Check the answer
5x + 30 = 180, so x = 30 and angle P = 3 × 30 + 10 = 100°. Check: R = 2 × 30 + 20 = 80° and 100° + 80° = 180°.
With a GDC (optional)
For a chord of length c that subtends an inscribed angle t, the radius comes from c = 2r sin t. Rearrange to r = c ÷ (2 sin t) and type it as c ÷ (2 × sin t) with the display in degrees (DEG), pressing equals once at the end.
The proofs in this lesson use no calculator. A scientific calculator does the same work for the one numerical question.
Check before you go on
1. The angles of four quadrilaterals are listed in order round each shape. For which one can you conclude that its four vertices lie on one circle?
Your turn
Start at Level 1 and go as far as you can. Type negative numbers with a minus sign, for example -7. A wrong answer gets a reason, not just a cross: read it, then try again.
Level 1 · Getting started
2. Triangle ABC has a right angle at B, with AB = 24 cm and BC = 10 cm. A circle passes through A, B and C. What is the radius of the circle?
3. Angle ABD = 37° and angle ACD = 37°, where B and C are on the same side of AD. Angle CAB = 64°. Taken in the order A, B, C, D, the points form a convex quadrilateral. Find angle CDB.
Level 2 · Building
4. PQRS is a cyclic quadrilateral with angle P = (4x + 15)°, angle Q = (2x + 40)° and angle R = (3x + 25)°. Find angle S.
5. ABCD is a cyclic quadrilateral. The side AB is extended to a point E. Angle ABC = 107°. Find angle ADC.
6. The mouth of a harbour is a gap AB of 60 m between two piers. A surveyor wants to stand where she sees AB at exactly 90°. How far must she be from the midpoint of AB, in metres?
Level 3 · Stretch
7. Two posts A and B are 20 m apart. A point P sees AB at an angle of 40°, and P, A and B lie on a circle. What is the radius of the circle, to 1 decimal place?
8. AB is a diameter of a circle and AB = 26 cm. The point C is on the circle with AC = 10 cm, and D is on the circle on the other side of AB with BD = 10 cm. Find the area of the quadrilateral ACBD.
Something to think about
In the first figure P sees AB at 50°. Describe every point on that side of AB that sees AB at exactly 56°. Is it on a circle through A and B that is bigger or smaller than the first one, and by roughly how much?
One way to think about it
An arc of another circle through A and B. In the isosceles triangle from the centre, half of AB = r sin t, so r = (AB ÷ 2) ÷ sin t. A bigger angle has a bigger sine, so the circle is smaller: its radius is sin 50° ÷ sin 56° ≈ 0.766 ÷ 0.829 ≈ 0.92 of the first.
Exit ticket
9. Points C and D are on opposite sides of the line AB. Angle ACB = 55° and angle ADB = 125°. What can you conclude?