• Quick retrieval from earlier learning (Circle theorems: angle at the centre and angle in a semicircle; Circle theorems: cyclic quadrilaterals and angles in the same segment). Students answer on mini whiteboards.
• Show answers together and note who needs support in the main task.

1. O is the centre of a circle. A, B and C are on the circle, with C on the major arc. ∠ACB = 25°. Find ∠AOB. → 50Prompt: Both angles stand on the same arc AB.
2. ABCD is a cyclic quadrilateral. ∠DAB = 122°. Find ∠BCD. → 58Prompt: Are the two angles opposite each other?
3. O is the centre. ∠AOB = 92° and C is on the major arc. Find ∠ACB. → 46Prompt: Angle at centre = 2 × angle at circumference.

• Show a photo of a building, a tile pattern or a road sign and ask what shapes and angles students can spot.
• Share the objective: Name the circle theorem that justifies a step. Read the success criteria together.

• Work through the example out loud, saying why you do each step, not just what you do.
• Example: O is the centre of the circle. Which theorem justifies this step? ∠OTP = 90°, where PT touches the circle at T
• Step 1: A tangent is perpendicular to the radius
• Answer: A tangent is perpendicular to the radius
• Pause after each step and ask a student to say what comes next.

• Do these together. Students try each step on whiteboards before you reveal it.
• Use the hints as prompts rather than giving the answer.

1. O is the centre of the circle. Which theorem justifies this step? ∠ABC + ∠ADC = 180° in quadrilateral ABCD drawn inside the circle (Tangents from an outside point are equal in length / Opposite angles of a cyclic quadrilateral add up to 180° / A tangent is perpendicular to the radius / Angles in the same segment are equal) → Opposite angles of a cyclic quadrilateral add up to 180°Prompt: Look at where the angle's vertex is: at the centre, on the circle, or outside it.
2. O is the centre of the circle. Which theorem justifies this step? AB is a diameter, so ∠ACB = 90° (Opposite angles of a cyclic quadrilateral add up to 180° / The angle in a semicircle is 90° / A tangent is perpendicular to the radius / Tangents from an outside point are equal in length) → The angle in a semicircle is 90°Prompt: Look at where the angle's vertex is: at the centre, on the circle, or outside it.

• Students work on the student sheet. Circulate and check the first two answers of every student early.
• Students who finish: write a question of their own for a partner, with the answer.

The 6 questions are on the student sheet below, with answers in the key.

• Two questions on their own. Sort answers into got it / nearly / not yet to plan the next lesson.

1. O is the centre of the circle. Which theorem justifies this step? PA = PB, where PA and PB touch the circle (The angle in a semicircle is 90° / The angle at the centre is twice the angle at the circumference / Tangents from an outside point are equal in length / A tangent is perpendicular to the radius) → Tangents from an outside point are equal in lengthPrompt: Look at where the angle's vertex is: at the centre, on the circle, or outside it.
2. O is the centre of the circle. Which theorem justifies this step? ∠ADB = ∠ACB, both standing on chord AB on the same side (Tangents from an outside point are equal in length / Opposite angles of a cyclic quadrilateral add up to 180° / The angle at the centre is twice the angle at the circumference / Angles in the same segment are equal) → Angles in the same segment are equalPrompt: Look at where the angle's vertex is: at the centre, on the circle, or outside it.

Free online practice with hints and printable worksheets for this skill: eurekami.com/math/grade-10.