Spatial Sense

Using think-pair-share activities, learners discuss the various circle theorem statements. Encourage students to have a decision-making role related to classroom activities and rules.
Example: Circle Theorems Statements
- The angle subtended by a chord at the centre is twice the angle subtended by it at the circumference.
- The angle subtended by the diameter at the circumference is a right angle.
- The angles subtended at the circumference by the same arc are equal.
- Two equal chords subtend equal angles at the centre of the circle.
- If the angles subtended by two chords at the centre are equal, then the two chords are equal.
- The opposite angles in a cyclic quadrilateral are supplementary.
- The angle between the radius and the tangent at the point of contact is 90 degrees.
Think-pair-share activities: In pairs, discuss the various circle theorem proofs.
Example: Circle Theorems Proofs
Theorem 1: The angle subtended by a chord at the centre is twice the angle subtended by it at the circumference.
Proof: Consider the following circle, in which an arc (or segment) AB subtends ∠AOB at the centre O and ∠ACB at a point C on the circumference. We have to prove that ∠AOB = 2 × ∠ACB. Draw a line segment through O and C, and let it intersect the circle again at point D, as shown.

There are two triangles formed: ΔOAC and ΔOBC. So, we make the following observations. In ΔOAC, ∠OAC = ∠OCA because OA = OC (OA and OC being the radii. Angles opposite to equal sides are equal). In ΔOBC, ∠OBC = ∠OCB because OB = OC (OB and OC being the radii. Angles opposite to equal sides are equal). Hence, using the exterior angle theorem, we get, ∠AOD= 2×∠ACO ⋯ (1) ∠DOB= 2×∠OCB ⋯ (2) Add equations (1) and (2): ∠AOD+ ∠DOB= 2× (∠ACO+ ∠OCB) ⇒ ∠AOB= 2× ∠ACB
Theorem 2: The angle subtended by the diameter at the circumference is a right angle.
Proof: Consider the figure below, where AB is the diameter of the circle. We need to prove that ∠ACB= 90°

Using theorem 1, 'The angle subtended by a chord at the centre is twice the angle subtended by it at the circumference.' We have ∠AOB= 2× ∠ACB. Now, ∠AOB = 180° as AB is a straight line (diameter). So, 2 × ∠ACB = 180° which implies ∠ACB = 90°.
Theorem 3: The angles subtended at the circumference by the same arc are equal.
Proof: Consider the following figure, which shows an arc AB subtending angles ACB and ADB at two arbitrary points, C and D, on the circumference. O is the centre of the circle. 

We need to prove that ∠ACB= ∠ADB.
Using the circle theorem, 'The angle subtended by a chord at the centre is twice the angle subtended by it at the circumference.' we have that ∠ACB= 1/2× ∠AOB ⋯ (1) ∠ADB= 1/2× ∠AOB ⋯ (2) From equations (1) and (2), we get ∠ACB= ∠ADB. Since angles ACB and ADB are arbitrary angles, therefore, the result is true for all angles subtended by the same arc.
Theorem 4: Two equal chords subtend equal angles at the centre of the circle.
Proof: Consider a circle given below with centre O and two chords AB and CD, such that AB = CD. Now, we need to prove ∠AOB = ∠COD. 

In triangles AOB and COD, we have OA = OC (Radii) OB = OD (Radii) AB = CD (Given)
So, triangles AOB and COD are congruent by SSS congruence rule. So, we have ∠AOB = ∠COD (Corresponding parts of congruent triangles).
Theorem 5: If the angles subtended by two chords at the centre are equal, then the two chords are equal.
Proof: Consider a circle given below with centre O and two chords, AB and CD, such that ∠AOB = ∠COD. Now, we need to prove AB = CD. In triangles AOB and COD, we have OA = OC (Radii) OB = OD (Radii) ∠AOB = ∠COD (Given) So, triangles AOB and COD are congruent by SAS congruence rule. So, we have AB = CD (Corresponding parts of congruent triangles).

Group discussions: In groups, task learners to apply the various circle theorems to solve some examples. Encourage learners to show respect to individuals of different backgrounds in their groups as they solve real-life problems on circle theorems.
Example 1: Consider a circle with Centre O given below. Find the value of x using circle theorems. 

Solution: We are given a circle with a centre O. Sine OS, and OT are radii, OS = OT. Using the circle theorem 'The angle between the radius and the tangent at the point of contact is 90 degrees', we have ∠OTP = 90°. In triangle OTP, using the angle sum theorem, we have ∠TOP + ∠OTP + ∠OPT = 180° ⇒ ∠TOP + 90° + 32° = 180° ⇒ ∠TOP = 180° - (90° + 32°) = 58° Since OS = OT ⇒ ∠OSP = ∠OTP = x (because angles opposite to equal sides are equal). Using the exterior angle theorem, we have ∠OSP + ∠OTP = ∠TOP ⇒ x + x = 58° ⇒ 2x = 58° ⇒ x = 29°
Answer: x = 29°
Example 2: Consider the circle given below with centre O. Find the angle x using the circle theorems. 

Solution: Using the circle theorem 'The angle subtended by the diameter at the circumference is a right angle', we have ∠ABC = 90°. So, using the triangle sum theorem, ∠BAC + ∠ACB + ∠ABC = 180° ⇒ x + 55° + 90° = 180° ⇒ x + 145° = 180° ⇒ x = 180° - 145° = 35° Answer: x = 35
Teaching and Learning Resources:
- * Mathematical sets. Graph sheet. * Technology tools such as computers, mobile phones, etc. * Computer software applications like GeoGebra.
Assessment (3.3.1.AS.2). The document marks these depth-of-knowledge levels for this indicator: Level 1 Recall; Level 2 Skills of conceptual understanding; Level 3 Strategic reasoning.