Lesson Notes By Weeks and Term v5 - Grade 11

Lesson Video

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Performance objectives

• Prove and apply the Angle at the Centre theorem.
• Solve problems with angles in the same segment.
• Prove and apply the theorem about opposite angles of a cyclic quadrilateral.
• Apply the exterior angle theorem for cyclic quadrilaterals.
• Justify steps in geometric proofs.

Lesson summary

This week, we delve deeper into the fascinating world of Euclidean Geometry, specifically focusing on theorems related to circles. Understanding circle geometry is not just about memorizing theorems; it's about developing your logical reasoning and problem-solving skills – crucial assets in various fields, from engineering and architecture to coding and even everyday decision-making. Think about how circles are used in the design of roundabouts to optimise traffic flow, in the construction of bridges to distribute weight effectively, or even in calculating the surface area when deciding how much paint is needed to paint a circular rondavel.

Lesson notes

2.1 Angle at the Centre Theorem Theorem: The angle subtended by a chord at the centre of a circle is twice the angle subtended by the same chord at the circumference.

Explanation: Imagine a circle with centre O. Pick two points on the circumference, A and B. The line segment connecting A and B is the chord AB. The angle AOB is the angle subtended by the chord AB at the centre O. Now, pick another point C on the circumference (on the same side of AB as the centre). The angle ACB is the angle subtended by the chord AB at the circumference. The Angle at the Centre Theorem states that angle AOB = 2 * angle ACB. Why is this true? Consider the line OC. It divides the triangle AOB. Also consider the triangle AOC. Since OA and OC are radii, the triangle AOC is isosceles.

Therefore angle OAC = angle OCA. If we let angle OAC = x, then angle AOC = 180 – 2x. Similarly, in triangle BOC, if we let angle OBC = y, then angle BOC = 180 – 2y. Then angle AOB = 360 - (180 -2x) - (180 - 2y) = 2x + 2y = 2(x + y). But x + y = angle AC

B. Hence angle AOB = 2 * angle AC

B. Example 1: In circle O, angle ACB = 35°. Calculate the size of angle AO

B. Solution: Angle AOB = 2 angle ACB (Angle at centre = 2 angle at circumference) Angle AOB = 2 * 35° Angle AOB = 70° 2.2 Angles in the Same Segment Theorem: Angles subtended by a chord at the circumference of a circle, on the same side of the chord, are equal.

Explanation: Again, consider a circle with centre O and chord AB. This time, pick two points C and D on the circumference, both on the same side of chord AB. Then angle ACB = angle ADB. These angles are said to be in the same segment. Why is this true? Both angles ACB and ADB are half of the angle subtended at the centre. So angle ACB = ½ angle AOB and angle ADB = ½ angle AO

B. Example 2: In circle O, angle APB = 48°. Calculate the size of angle AQ

B. Solution: Angle AQB = angle APB (Angles in the same segment) Angle AQB = 48° 2.3 Cyclic Quadrilaterals Definition: A cyclic quadrilateral is a quadrilateral whose vertices all lie on the circumference of a circle.

Theorem 1: The opposite angles of a cyclic quadrilateral are supplementary (add up to 180°).

Explanation: If ABCD is a cyclic quadrilateral, then angle A + angle C = 180° and angle B + angle D = 180°. Why is this true? Consider the angles subtended at the centre by chord CD, namely reflex angle COD. Then, angle A = ½ reflex angle COD. Similarly, angle C = ½ angle BOD (where B and D are the other two vertices of the cyclic quadrilateral. Hence A + C = ½ (360) =

1

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0. Theorem 2: The exterior angle of a cyclic quadrilateral is equal to the interior opposite angle.

Explanation: If we extend one side of a cyclic quadrilateral (say, side BC to point E), then angle DCE (the exterior angle) is equal to angle A (the interior opposite angle). Why is this true? Angle DCE + angle BCD = 180 (Angles on a straight line) and angle A + angle BCD = 180 (opposite angles of cyclic quad). Thus angle DCE = angle

A. Example 3: ABCD is a cyclic quadrilateral. Angle ABC = 105°. Calculate the size of angle AD

C. Solution: Angle ADC + angle ABC = 180° (Opposite angles of cyclic quad) Angle ADC + 105° = 180° Angle ADC = 180° - 105° Angle ADC = 75° Example 4: In cyclic quadrilateral PQRS, side PQ is extended to

T. If angle R = 72°, calculate the size of angle TQ

P. Solution: Angle TQP = angle R (Exterior angle of cyclic quad = interior opposite angle) Angle TQP = 72° Guided Practice (With Solutions)

Question 1: In circle O, O is the centre and angle AOB = 110°. Calculate the size of angle AC

B. Solution: Angle AOB = 2 angle ACB (Angle at centre = 2 angle at circumference) 110° = 2 * angle ACB Angle ACB = 110° / 2 Angle ACB = 55°

Commentary: This question directly applies the Angle at the Centre Theorem. We identify the angle at the centre and the angle at the circumference subtended by the same chord (AB) and then use the theorem's relationship.

Question 2: A, B, C, and D are points on the circumference of a circle. If angle BAC = 40° and angle CAD = 25°, calculate the size of angle BD

C. Solution: Angle BCD = angle BAC + angle CAD = 40 + 25 =

6

5. Angle BDC = angle BAC (angles in same segment). Angle BDC = 40°.

Commentary: Here, we first have to recognize that angle BAD consists of two smaller angles. The angle BDC is in the same segment as BAC so they are equal.

Question 3: PQRS is a cyclic quadrilateral. If angle P = x + 20° and angle R = 2x - 10°, calculate the value of x.

Solution: Angle P + angle R = 180° (Opposite angles of cyclic quad are supplementary) (x + 20°) + (2x - 10°) = 180° 3x + 10° = 180° 3x = 170° x = 170° / 3 x ≈ 56.67°

Commentary: This question involves setting up an equation using the property of cyclic quadrilaterals, where opposite angles sum to 180°. Solving the equation leads to the value of the unknown variable.

Question 4: In cyclic quadrilateral ABCD, BC is extended to E. If angle DCE = 82°, determine the size of angle BAD.