1. Day 122 – Angle properties of a circle 1

2. Introduction
Lines drawn on a circle have different names depending on the endpoints of the line. A line drawn on a circle can either be a diameter, a chord, a radius, a secant or a tangent. Similarly, angles on a circle have different identities depending on the lines which make them. Some of these angles include the central, inscribed and circumscribed angles. These parts of a circle have different relationships. In this lesson, we will identify and describe the relationships among inscribed angles, radii and chords.

3. Vocabulary
Chord of a circle This is a line segment with its endpoints lying on the circumference of a circle. Inscribed angle This is an angle on a circle with its vertex being a meeting point of two chords with a common end point. Radius of a circle This is the distance from the center of a circle to its circumference.

4. Consider the figure below.
Chord of a circle
If a line segment is drawn on a circle such that its endpoints lie on the circumference of the circle, that line segment is called the chord of a circle.
Consider the figure below.
The line segments AB and CD are chords of the circle. A B C D

5. Example 1 Identify the name of the line segment KL. Explain your answer. Solution Chord Its endpoints lie on the circumference. K L

6. If chord is perpendicular to the radius of a circle, then the radius passes through the midpoint of that chord. Consider the figure below. OM is the radius of the circle which intersects chord JL at right angle. 𝐽𝐾=𝐾𝐿. O K J L M

7. Proof We want to prove that given that OM⊥𝐽𝐿 then, 𝐽𝐾= 𝐾𝐿
Proof We want to prove that given that OM⊥𝐽𝐿 then, 𝐽𝐾= 𝐾𝐿. Since OJ and OL are radii of the same circle, then OJ = OL. ∠𝐽𝐾𝑂=∠𝐿𝐾𝑂= 90 ° OK is shared by ∆𝐽𝑂𝐾 and ∆𝐾𝑂𝐿. By S.A.S postulate ∆𝐽𝑂𝐾≅∆𝐾𝑂𝐿. JK and KL are corresponding sides thus 𝐽𝐾=𝐾𝐿. O K J L M

8. Any line segment that passes through the center of the circle and perpendicular to a chord, bisects the chord. Conversely, a perpendicular bisector to a chord must pass through the center of the circle.

9. If two chords have equal distance from the center of the circle, then they are equal in length. Since the two chords are equidistance from the center of the circle, then 𝐴𝐷=𝐵𝐶. A B C D 𝑥 O

10. Proof We want to prove that 𝐴𝐷=𝐶𝐵 if 𝑂𝑃=𝑂𝑀
Proof We want to prove that 𝐴𝐷=𝐶𝐵 if 𝑂𝑃=𝑂𝑀. 𝑂𝐶=𝑂𝐴 (radii of the same circle) 𝑂𝑃=𝑂𝑀( given) The radius are perpendicular to the chord by the previous results. By Hypotenuse-Leg postulate ∆𝐶𝑃𝑂≅𝐴𝑀𝑂 CP and AM are corresponding sides and therefore equal. A B C D 𝑀 𝑃 O

11. Any line segment that passes through the center of the circle and perpendicular to a chord, bisects the chord. Therefore,𝐶𝑃=𝑃𝑀=𝐴𝑀=𝑀𝐷 𝐶𝑃+𝑃𝐵=𝐴𝑀+𝑀𝐷 But 𝐶𝑃+𝑃𝐵=CB and 𝐴𝑀=𝑀𝐷=𝐴𝐷 thus, 𝐶𝐵=𝐴𝐷

12. If two chord are equal, then they subtend the equal angles at the center. Proof We want prove that ∠𝑀𝑂𝑁=∠𝑃𝑂𝑄 Since 𝑁𝑂, 𝑀𝑂, 𝑃𝑂 and 𝑄𝑂 are radii of the same circle then, 𝑁𝑂=𝑀𝑂=𝑃𝑂=𝑄𝑂. By S.S.S postulate, ∆𝑀𝑂𝑁≅∆𝑃𝑂𝑄 M N O P Q

13. ∠𝑀𝑂𝑁 and ∠𝑃𝑂𝑄 are corresponding angles. Therefore, ∠𝑀𝑂𝑁=∠𝑃𝑂𝑄
∠𝑀𝑂𝑁 and ∠𝑃𝑂𝑄 are corresponding angles. Therefore, ∠𝑀𝑂𝑁=∠𝑃𝑂𝑄. Example 2 Find the value of 𝑥 if ∠𝐴𝑂𝐷= 4𝑥+42 ° and ∠𝐵𝑂𝐶= 5𝑥+12 ° C B O A D

14. Solution Since ∠𝐴𝑂𝐷 and ∠𝐵𝑂𝐶 are two central angles subtended by equal chords then, ∠𝐴𝑂𝐷=∠𝐵𝑂𝐶 4𝑥+42=5𝑥+12 5𝑥−4𝑥=42−12 𝑥=30

15. Inscribed angle If two chords meet at a common point making an angle, that angle is referred as an inscribed angle. AB and BC are two chords which meet at point B making an angle. ∠𝐵 is an inscribed angle. B A C

16. The central angle is twice the size of the inscribed angle subtended by the same arc. ∠𝐴𝑂𝐶=2∠𝐴𝐵𝐶 Proof ∠𝐴𝐵𝐶=𝜃+𝛼 𝑂𝐴=𝑂𝐵=𝑂𝐶 (radii of the same circle) Triangles AOB and BOC are isosceles triangles thus base angles are equal in each triangle. B A C 𝜃 𝛼 O

17. ∠𝐴𝑂𝐵=180−2𝜃 (angle sum of a triangle is 180 ° ) Similarly, ∠𝐵𝑂𝐶=180−2𝛼 ∠𝐵𝑂𝐶+∠𝐴𝑂B+∠𝐴𝑂𝐶= 360 ° (angle sum at the center) (180−2𝛼)+(180−2𝜃)+∠𝐴𝑂𝐶= 360 ° ∠𝐴𝑂𝐶=360−360+ 2𝜃+2𝛼 ∠𝐴𝑂𝐶=2(𝜃+𝛼) But ∠𝐴𝐵𝐶=𝜃+𝛼 Therefore, ∠𝐴𝑂𝐶=2 ∠𝐴𝐵𝐶

18. We will also proof that the central angle(reflex angle) is twice the inscribed angle subtended by the same arc Consider the figure below. We will show that the reflex ∠𝐴𝑂𝐶=2∠𝐴𝐵𝐶. Let ∠𝐴𝑂𝐵=𝑏 and ∠𝐵𝑂𝐶=𝑎. 𝑂𝐴=𝑂𝐵=𝑂𝐶 (radii of the same circle are equal) In ∆𝐵𝑂𝐶,∠𝑂𝐶𝐵=∠𝑂𝐵𝐶= −𝑎 =90− 𝑎 2 since the sum of angles of a triangle is 180 ° . B A C 𝑏 𝑎 O

19. In ∆𝐵𝑂𝐴,∠𝑂𝐴𝐵=∠𝑂𝐵𝐴= −𝑏 =90− 𝑏 2 since the sum of angles of a triangle is 180 ° . ∠𝐴𝐵𝐶= ∠𝑂𝐵𝐶+∠𝑂𝐵𝐴 =90− 𝑎 2 +90− 𝑏 2 =180− 𝑎 2 − 𝑏 2 Since the angle sum at the center of a circle is 360 ° , reflex ∠𝐴𝑂𝐶=360−𝑎−𝑏 = 1 2 (180− 𝑎 2 − 𝑏 2 ) But 180− 𝑎 2 − 𝑏 2 = ∠𝐴𝐵𝐶 Therefore reflex ∠𝐴𝑂𝐶=2∠𝐴𝐵𝐶

20. Example 3 Find the size of ∠𝑁𝑀𝑂. P is the center of the circle
Example 3 Find the size of ∠𝑁𝑀𝑂. P is the center of the circle. Solution ∠𝑁𝑃𝑂=2∠𝑁𝑀𝑂 89 ° =2∠𝑁𝑀𝑂 ∠𝑁𝑀𝑂= 89 2 = 44.5 ° M N O P
89 °

21. Inscribed angles subtended on one segment by a common chord are equal
Inscribed angles subtended on one segment by a common chord are equal. ∠𝑃=∠𝑄 since they share a common chord MN. M N P Q

22. ∠𝑃=∠𝑄 Proof ∠𝑀𝑂𝑁 is a central angle subtended by arc MN
∠𝑃=∠𝑄 Proof ∠𝑀𝑂𝑁 is a central angle subtended by arc MN. ∠𝑀𝑃𝑁 is an inscribed angle subtended by arc MN. Therefore ∠𝑀𝑃𝑁= 1 2 ∠𝑀𝑂𝑁 ∠𝑀𝑄𝑁 is an inscribed angle subtended by arc MN. Therefore ∠𝑀𝑄𝑁= 1 2 ∠𝑀𝑂𝑁 Thus Therefore ∠𝑀𝑃𝑁=∠𝑀𝑄𝑁 M N P Q O

23. M N P Q K J
Inscribed angles subtended by equal chords are equal. If 𝑀𝑁=𝐽𝐾 then ∠𝑀𝑄𝑁=∠𝐾𝑃𝐽.

24. 𝑀𝑁=𝐽𝐾 Proof Since 𝑀𝑁 and 𝐽𝐾 are equal chords, then they subtend equal angles at the center therefore ∠𝑀𝑂𝑁=∠𝐽𝑂𝐾 ∠𝑀𝑄𝑁= 1 2 ∠𝑀𝑂𝑁= 1 2 ∠𝐽𝑂𝐾 (Inscribed angle is half the central angle) ∠𝐾𝑃𝐽= 1 2 ∠𝐽𝑂𝐾 (inscribed angle is half the central angle) ∠𝐾𝑃𝐽=∠𝑀𝑄𝑁 M N P Q K J O

25. homework Find the size of obtuse ∠𝐽𝑂𝐿 in the figure below. K L 65 ° OJ K O
65 °

26. Answers to homework
130 °

27. THE END