At a glance. One lesson of about 60 minutes. It assumes the circle theorems from Standard. Students state and prove the converses of the semicircle, same-segment and cyclic quadrilateral theorems, and use them to show points are concyclic.
Calculator: GDC optional. A calculator is used for one sine, so the label is GDC optional. The proofs are done by hand; a scientific calculator does the same work.
What students will learn
Know
- An angle of 90° at C puts C on the circle with diameter AB.
- Equal angles on a chord from the same side put the points on one circle.
- Opposite angles adding to 180° make a quadrilateral cyclic.
Understand
- A theorem and its converse are different statements and each needs a proof.
- A point inside a circle sees a chord at a bigger angle, outside at a smaller one, so only the circle gives the exact angle.
Do
- Prove a converse by construction or by contradiction.
- Use a converse to show points are concyclic and then apply a theorem.
- Find the radius from a chord and an inscribed angle.
MYP assessment criteria
Wording paraphrased. What this lesson gives you evidence for, and what to look for:
- Criterion A. Knowing and understanding: stating and using the converses of the circle theorems (questions 1 to 9).
- Criterion B. Investigating patterns: the first figure. Students describe how the angle changes inside, on and outside the circle.
- Criterion C. Communicating: proofs with every reason written, including the case of a point inside and outside (worked examples 1 to 3).
- Criterion D. Applying mathematics in real-life contexts: the harbour mouth and the surveyor's position.
The one thing to do with the figure
Make students predict, then reveal. Before the first figure, ask: where can the surveyor stand? Take three guesses on the board. Then move P round the circle at offset 0 and then at 10% in and out.
In the second figure, ask which of the three distances MA, MB and MC they would have guessed to be equal, and why the third is not obvious.
A possible order
| 5 min | Warm-up from the student page, in silence, then answers. |
|---|---|
| 5 min | Hook: the harbour mouth. Collect votes and reasons. |
| 12 min | First figure; key idea and the contradiction proof; settle the hook; the warning card. |
| 10 min | Second figure; worked examples 1 to 3; pairs finish the faded example. |
| 5 min | Question 1, the check. If many choose B or D, return to which angles are opposite. |
| 18 min | Your turn, Levels 1 to 3, at each student's own pace. |
| 5 min | Something to think about, then the exit ticket, question 9. |
The answers
| Warm-up | 42°; 109°; 75°. |
|---|---|
| Worked examples | The proof by rectangle; 27°; the proof by contradiction. |
| Faded example | 100°. |
| Something to think about | An arc of a smaller circle through A and B, about 0.92 of the radius. |
| 1. Using the converse of the cyclic quadrilateral theorem | A. B and D add neighbouring angles; C trusts two right angles. |
| 2. Radius of the circle through a right-angled triangle | 13 cm. AC = 26, halved. |
| 3. Concyclic first, then equal angles | 64°. CDB = CAB. |
| 4. A cyclic condition as an equation | 100°. x = 20, Q = 80. |
| 5. Exterior angle of a cyclic quadrilateral | 68°. ABC = 112, then 180 − 112. |
| 6. Locating a viewing point with a right angle | 30 m. Radius of the circle with diameter AB. |
| 7. Radius from a chord and an inscribed angle | 15.6 m. 10 ÷ sin 40°. |
| 8. Area of a quadrilateral inscribed with a diameter | 240 cm². 2 × 120. |
| 9. Concyclic from opposite sides | C. A uses the same-side test across AB; B wants a diameter; D thinks more angles are needed. |
What each wrong answer tells you
| They give | What it usually means | What to do |
|---|---|---|
| B, C or D in question 1 | Added neighbouring angles instead of opposite ones, or took two right angles as enough. | Label the vertices in order and join each to the one opposite before adding. |
| 26 or 17 in question 2 | Gave the diameter, or averaged the legs. | Mark the centre on the hypotenuse and say what the radius is. |
| 37 or 79 in question 3 | Gave a given angle, or the angle where the diagonals cross. | First say why the points are concyclic, then shade the chord CB. |
| 80 or 95 in question 4 | Gave Q, or gave P. | Underline what is asked, angle S, and mark the angle opposite it. |
| 112 or 34 in question 5 | Gave the interior angle ABC, or halved. | Mark the straight line ABE and the two angles on it at B. |
| 60 or 42.4 in question 6 | Gave the whole gap, or used 45°. | Draw the circle with diameter AB and place P on it. |
| 10.2 or 29.2 in question 7 | Used the whole centre angle, or halved the inscribed angle. | Draw the centre angle, then the isosceles triangle, and halve. |
| 120 or 130 in question 8 | Gave one triangle, or took AB as a base with AC as height. | Ask which angle is the right angle, and then which are base and height. |
| A, B or D in question 9 | Used the equal-angle test across AB, asked for a diameter, or thought more angles were needed. | Ask: which side of AB is each point on, and which converse fits that picture? |
Questions worth asking
- Why is a theorem not the same as its converse? Listen for “one says what follows from being on the circle; the other says that a condition forces the points onto it”.
- Why does the proof by contradiction need Q inside and outside? Listen for “both cases make the angle different”.
- Equal angles on a chord are not enough. What else must you check, and why? Listen for “the other side gives 180° minus the angle”.
- How does the rectangle in worked example 1 put C on the circle? Listen for “diagonals of a rectangle are equal and bisect each other”.
Differentiation
A way in for anyone stuck
- Give the three converses on a card with a small diagram each.
- Provide a sentence frame for the contradiction: Suppose Q is not on the circle. Then Q is …, so the angle is …, which is a contradiction.
- Restrict to questions 1 to 4 at first, using the given figures.
Stretch
- Prove that the locus of points P on one side of AB with angle APB = t is an arc of a circle, and find its centre.
- Show that a cyclic trapezium must be isosceles, with its two non-parallel sides equal.
Students learning in English as an additional language
“Converse” means the statement the other way round: “if A then B” becomes “if B then A”. “Concyclic” means on the same circle. “Contradiction” means two statements that cannot both be true. Use a pair of everyday if-then sentences first.
| Word | Plain meaning | Example |
|---|---|---|
| converse | the statement turned round: if B then A instead of if A then B | if it is a square, it has four sides; the converse is false |
| concyclic | lying on one circle | four posts round a circular pond |
| contradiction | a result that cannot be true at the same time as what you assumed | assume it is raining and dry at once |
| subtend | to make an angle at a point from the ends of a line | the harbour mouth subtends an angle at the surveyor |
| diameter | a chord through the centre; the longest chord | the width of a round table |
Sentence starters for explaining aloud or in writing:
- Suppose the point is not on the circle. Then it is …
- The points are concyclic because … on the same side of …
- The converse says … and the theorem says …
Two things not to say
“It is obvious the other way round too.” Prove it.
“Equal angles mean the same circle.” Same side and same chord are needed.
Questions to set
Extra practice in three tiers, including one question from earlier work for spacing. Every answer has been checked.
1Fluency
The idea on its own, with friendly numbers. Set these first.
- ConverseA triangle has sides 9, 12 and 15. Which side is a diameter of its circle?
The side 15. - Earlier work (Year 10)The angle at the centre is 150°. Find the angle at the circumference on the same arc.
75°.
2In context
The same idea inside a situation, where the first job is deciding what to calculate.
- RadiusA right-angled triangle has legs 5 and 12. Find the radius of its circle.
6.5. - Cyclic testA quadrilateral has angles 75°, 110°, 105° and 70° in order round it. Is it cyclic?
Yes: 75 + 105 = 180 and 110 + 70 = 180.
3Challenge
Reasoning, generalizing and working backwards.
- ReasoningExplain why a parallelogram that is cyclic must be a rectangle.
Opposite angles are equal and add to 180°, so each is 90°. - OpenA point P above the line AB sees AB at 50°. Describe where a point below the line must be to see AB at 130°.
On the arc of the same circle that lies below AB, the minor arc.
Practicalities
Works on a phone or a shared screen. Nothing a student types is saved or sent anywhere.
Both figures work with the keyboard (tab to a slider, then the arrow keys) and by touch. “Show me the steps” gives the same states in words.
No device? Two pins and a loop of string on a board trace the arc of a circle through two fixed points.