Lesson Note on Circle Theorems: Proofs of Major Theorems for SSS 2

Class: Senior Secondary School 2 (SS2 / SSS2)
Term: First Term
Week: 3
Age: 16 years
Duration: 45 minutes
Subject: General Mathematics
Curriculum Theme: Circle Geometry
Previous Lesson: .
Topic: CIRCLE THEOREMS
Subject Matter: Proof of the theorem: the angle which an arc suspends at centre is twice the angle subtended at the circumference, proof of the theorem: the angles in same segment are equal, proof of the theorem: the angles in a semi-circle are one right angle, proof of the theorem: the opposite angles of a cyclic quadrilateral are supplementary, proof of the theorem: the tangent to a circle is perpendicular to the radius

Specific Objectives

By the end of the lesson, pupils should be able to:

Cognitive Domain:

• State the five basic circle theorems.
• Explain the conditions under which each theorem applies.
• Demonstrate the proof for each of the five theorems.

Affective Domain:

• Appreciate the logical reasoning involved in geometric proofs.
• Show interest in solving problems related to circle theorems.
• Participate actively in class discussions.

Psychomotor Domain:

• Construct accurate diagrams to illustrate each theorem.
• Apply the theorems to solve mathematical problems involving angles in circles.
• Write out clear, step-by-step proofs for each theorem.

Social Domain:

• Collaborate with peers to discuss proofs and solutions.
• Share ideas and strategies for solving circle theorem problems.

Reference Materials

The following resources were used in planning this lesson:

• Senior Secondary Schools Education Curriculum
• State Unified Scheme of Work
• New General Mathematics for Senior Secondary Schools 2

Instructional Materials

The teacher will teach this lesson with the aid of:

• Models of circle theorems
• Whiteboard/chalkboard
• Markers/chalk
• Ruler and compass
• Protractor
• Geometric diagrams illustrating circles and angles

Rationale for the Lesson

This lesson helps pupils understand the fundamental properties of angles and lines in circles. It enables them to apply these properties to solve various geometric problems and strengthens their logical reasoning skills in mathematics.

Prerequisite/Previous Knowledge

Pupils have basic knowledge of circles, their parts (radius, diameter, chord, arc, segment), types of angles (acute, obtuse, right, reflex), properties of triangles (isosceles, exterior angles), and basic Euclidean geometry proofs.

Lesson Content/Board Summary

CIRCLE THEOREMS

Theorem 1: The angle which an arc subtends at the centre is twice the angle it subtends at any point on the remaining part of the circumference.

Proof of Theorem 1

Given: A circle with centre O and an arc PQ subtending ∠POQ at the centre and ∠PAQ at the circumference.

Required to Prove: ∠POQ = 2∠PAQ.

Construction: Join AO and produce it to a point B.

Proof:

• In ΔAPO, OA = OP (radii). Therefore, ΔAPO is an isosceles triangle.
• ∠OAP = ∠OPA (base angles of isosceles triangle).
• ∠POB = ∠OAP + ∠OPA (exterior angle of ΔAPO).
• ∠POB = 2∠OAP (since ∠OAP = ∠OPA).
• Similarly, in ΔAQO, OA = OQ (radii). Therefore, ΔAQO is an isosceles triangle.
• ∠OAQ = ∠OQA.
• ∠QOB = ∠OAQ + ∠OQA (exterior angle of ΔAQO).
• ∠QOB = 2∠OAQ.
• Adding the two exterior angles: ∠POB + ∠QOB = 2∠OAP + 2∠OAQ.
• ∠POQ = 2(∠OAP + ∠OAQ).
• ∠POQ = 2∠PAQ.

Thus, the angle at the centre is twice the angle at the circumference.

Theorem 2: Angles in the same segment are equal.

Proof of Theorem 2

Given: A circle with centre O and points P, Q, R, S on the circumference such that ∠PSQ and ∠PRQ are angles in the same segment, subtended by the arc PQ.

Required to Prove: ∠PSQ = ∠PRQ.

Construction: Join OP and OQ.

Proof:

• ∠POQ is the angle at the centre subtended by arc PQ.
• ∠PSQ is the angle at the circumference subtended by arc PQ.
• From Theorem 1, ∠POQ = 2∠PSQ.
• Also, ∠PRQ is the angle at the circumference subtended by arc PQ.
• From Theorem 1, ∠POQ = 2∠PRQ.
• Therefore, 2∠PSQ = 2∠PRQ.
• Hence, ∠PSQ = ∠PRQ.

Thus, angles in the same segment are equal.

Theorem 3: The angle in a semi-circle is a right angle (90°).

Proof of Theorem 3

Given: A circle with centre O and diameter AB. C is any point on the circumference.

Required to Prove: ∠ACB = 90°.

Proof:

• The angle subtended by the diameter AB at the centre O is ∠AOB, which is a straight angle.
• ∠AOB = 180°.
• ∠ACB is the angle at the circumference subtended by the same arc AB (which is a semi-circle).
• From Theorem 1, the angle at the centre is twice the angle at the circumference.
• So, ∠AOB = 2∠ACB.
• 180° = 2∠ACB.
• ∠ACB = 180° / 2 = 90°.

Thus, the angle in a semi-circle is a right angle.

Theorem 4: The opposite angles of a cyclic quadrilateral are supplementary (add up to 180°).

Proof of Theorem 4

Given: A cyclic quadrilateral ABCD inscribed in a circle with centre O.

Required to Prove: ∠A + ∠C = 180° and ∠B + ∠D = 180°.

Construction: Join OB and OD.

Proof:

• The reflex angle ∠BOD subtended by arc BCD at the centre is 2∠BAD (Theorem 1).
• The obtuse angle ∠BOD subtended by arc BAD at the centre is 2∠BCD (Theorem 1).
• The sum of angles at the centre is 360°. So, reflex ∠BOD + obtuse ∠BOD = 360°.
• Therefore, 2∠BAD + 2∠BCD = 360°.
• Dividing by 2, ∠BAD + ∠BCD = 180°.
• Similarly, by joining OA and OC, it can be proven that ∠ABC + ∠ADC = 180°.

Thus, the opposite angles of a cyclic quadrilateral are supplementary.

Theorem 5: The tangent to a circle is perpendicular to the radius at the point of contact.

Proof of Theorem 5

Given: A circle with centre O and a tangent line XY touching the circle at point P. OP is the radius.

Required to Prove: OP ⊥ XY (i.e., ∠OPX = 90°).

Construction: Take any point Q on XY other than P. Join OQ.

Proof:

• If Q is any point on the tangent line XY other than P, then Q must lie outside the circle.
• Therefore, OQ must be longer than the radius OP (OQ > OP).
• This means OP is the shortest distance from the centre O to the line XY.
• The shortest distance from a point to a line is the perpendicular distance.
• Hence, OP is perpendicular to XY.

Thus, the tangent to a circle is perpendicular to the radius at the point of contact.

Application of Circle Theorems

Example 1: In a circle with centre O, if ∠AOB = 110°, find ∠ACB where A, B, C are points on the circumference.

Solution:

• Step 1: Identify the angles. ∠AOB is the angle at the centre, and ∠ACB is the angle at the circumference, both subtended by arc AB.
• Step 2: Apply Theorem 1. Angle at centre = 2 × Angle at circumference.
• Step 3: Calculate ∠ACB. 110° = 2 × ∠ACB. Therefore, ∠ACB = 110° / 2 = 55°.

Example 2: A, B, C, D are points on the circumference of a circle such that ABCD is a cyclic quadrilateral. If ∠ABC = 100°, find ∠ADC.

Solution:

• Step 1: Identify the figure. ABCD is a cyclic quadrilateral.
• Step 2: Apply Theorem 4. Opposite angles of a cyclic quadrilateral are supplementary.
• Step 3: Calculate ∠ADC. ∠ABC + ∠ADC = 180°. So, 100° + ∠ADC = 180°. ∠ADC = 180° – 100° = 80°.

Teaching Methods/Instructional Techniques

Discussion, Lecture, Demonstration, Question and Answer, Visual Aids

Instructional Procedures

Step 1: Introduction

Time: 5 minutes
Teaching Skill: Set Induction
Teacher’s Activity: The teacher greets the pupils and reviews the format for proving Euclidean Geometry, asking pupils to recall terms like Given, Required to Prove, Construction, and Proof.
Pupils’ Activity: Pupils respond to greetings and participate in the revision by mentioning the format along with the teacher.
Learning Point: Pupils recall the foundational steps for geometric proofs.

Step 2: Presentation & Proof of Theorem 1

Time: 7 minutes
Teaching Skill: Explanation/Demonstration
Teacher’s Activity: The teacher presents Theorem 1 (angle at centre is twice angle at circumference), draws a clear diagram on the board, and leads pupils through its proof, explaining each step.
Pupils’ Activity: Pupils observe the diagram, listen to the explanation, and suggest reasons for conclusions during the proof process.
Learning Point: Pupils understand and can reproduce the proof for Theorem 1.

Step 3: Presentation & Proof of Theorem 2

Time: 7 minutes
Teaching Skill: Explanation/Demonstration
Teacher’s Activity: The teacher presents Theorem 2 (angles in the same segment are equal), draws a diagram, and guides pupils through its proof, relating it to Theorem 1.
Pupils’ Activity: Pupils follow the proof, ask questions for clarification, and note down key points.
Learning Point: Pupils understand and can reproduce the proof for Theorem 2.

Step 4: Presentation & Proof of Theorem 3

Time: 7 minutes
Teaching Skill: Explanation/Demonstration
Teacher’s Activity: The teacher presents Theorem 3 (angle in a semi-circle is 90°), illustrates it with a diagram, and explains its proof, again linking it to Theorem 1.
Pupils’ Activity: Pupils observe the demonstration, participate in discussions, and contribute ideas for the proof.
Learning Point: Pupils understand and can reproduce the proof for Theorem 3.

Step 5: Presentation & Proof of Theorem 4

Time: 6 minutes
Teaching Skill: Explanation/Demonstration
Teacher’s Activity: The teacher presents Theorem 4 (opposite angles of a cyclic quadrilateral are supplementary), draws a diagram, and guides pupils through its proof.
Pupils’ Activity: Pupils follow the proof, making sure to understand the construction and reasoning.
Learning Point: Pupils understand and can reproduce the proof for Theorem 4.

Step 6: Presentation & Proof of Theorem 5

Time: 6 minutes
Teaching Skill: Explanation/Demonstration
Teacher’s Activity