OF A CIRCLE 1 I NTRODUCTION Lines drawn on a circle have different - - PowerPoint PPT Presentation

DAY 122 – ANGLE PROPERTIES

OF A CIRCLE 1

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.

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.

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

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

K L

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. 𝐾𝐿 = 𝐿𝑀.

O K J L M

Proof We want to prove that given that OM ⊥ 𝐾𝑀 then, 𝐾𝐿 = 𝐿𝑀. Since OJ and OL are radii of the same circle, then OJ = OL. ∠𝐾𝐿𝑃 = ∠𝑀𝐿𝑃 = 90° OK is shared by ∆𝐾𝑃𝐿 and ∆𝐿𝑃𝑀. By S.A.S postulate ∆𝐾𝑃𝐿 ≅ ∆𝐿𝑃𝑀. JK and KL are corresponding sides thus 𝐾𝐿 = 𝐿𝑀.

O K J L M

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.

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

• f the circle, then 𝐵𝐸 = 𝐶𝐷.

A B C D 𝑦 𝑦 O

Proof We want to prove that 𝐵𝐸 = 𝐷𝐶 if 𝑃𝑄 = 𝑃𝑁. 𝑃𝐷 = 𝑃𝐵 (radii of the same circle) 𝑃𝑄 = 𝑃𝑁( given) The radius are perpendicular to the chord by the previous results. By Hypotenuse-Leg postulate ∆𝐷𝑄𝑃 ≅ 𝐵𝑁𝑃 CP and AM are corresponding sides and therefore equal.

A B C D 𝑁 𝑄 O

Any line segment that passes through the center of the circle and perpendicular to a chord, bisects the chord. Therefore,𝐷𝑄 = 𝑄𝑁 = 𝐵𝑁 = 𝑁𝐸 𝐷𝑄 + 𝑄𝐶 = 𝐵𝑁 + 𝑁𝐸 But 𝐷𝑄 + 𝑄𝐶 =CB and 𝐵𝑁 = 𝑁𝐸 = 𝐵𝐸 thus, 𝐷𝐶 = 𝐵𝐸

If two chord are equal, then they subtend the equal angles at the center. Proof We want prove that ∠𝑁𝑃𝑂 = ∠𝑄𝑃𝑅 Since 𝑂𝑃, 𝑁𝑃, 𝑄𝑃 and 𝑅𝑃 are radii of the same circle then, 𝑂𝑃 = 𝑁𝑃 = 𝑄𝑃 = 𝑅𝑃. By S.S.S postulate, ∆𝑁𝑃𝑂 ≅ ∆𝑄𝑃𝑅

M N O P Q

∠𝑁𝑃𝑂 and ∠𝑄𝑃𝑅 are corresponding angles. Therefore, ∠𝑁𝑃𝑂 = ∠𝑄𝑃𝑅. Example 2 Find the value of 𝑦 if ∠𝐵𝑃𝐸 = 4𝑦 + 42 ° and ∠𝐶𝑃𝐷 = 5𝑦 + 12 °

C B O A D

Solution Since ∠𝐵𝑃𝐸 and ∠𝐶𝑃𝐷 are two central angles subtended by equal chords then, ∠𝐵𝑃𝐸 = ∠𝐶𝑃𝐷 4𝑦 + 42 = 5𝑦 + 12 5𝑦 − 4𝑦 = 42 − 12 𝑦 = 30

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. ∠𝐶 is an inscribed angle.

B A C

The central angle is twice the size of the inscribed angle subtended by the same arc. ∠𝐵𝑃𝐷 = 2∠𝐵𝐶𝐷 Proof ∠𝐵𝐶𝐷 = 𝜄 + 𝛽 𝑃𝐵 = 𝑃𝐶 = 𝑃𝐷 (radii of the same circle) Triangles AOB and BOC are isosceles triangles thus base angles are equal in each triangle.

B A C 𝜄 𝜄 𝛽 𝛽 O

∠𝐵𝑃𝐶 = 180 − 2𝜄 (angle sum of a triangle is 180°) Similarly, ∠𝐶𝑃𝐷 = 180 − 2𝛽 ∠𝐶𝑃𝐷 + ∠𝐵𝑃B + ∠𝐵𝑃𝐷 = 360°(angle sum at the center) (180 − 2𝛽) + (180 − 2𝜄) + ∠𝐵𝑃𝐷 = 360° ∠𝐵𝑃𝐷 = 360 − 360 + 2𝜄 + 2𝛽 ∠𝐵𝑃𝐷 = 2(𝜄 + 𝛽) But ∠𝐵𝐶𝐷 = 𝜄 + 𝛽 Therefore, ∠𝐵𝑃𝐷 = 2 ∠𝐵𝐶𝐷

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 ∠𝐵𝑃𝐷 = 2∠𝐵𝐶𝐷. Let ∠𝐵𝑃𝐶 = 𝑐 and ∠𝐶𝑃𝐷 = 𝑏. 𝑃𝐵 = 𝑃𝐶 = 𝑃𝐷 (radii of the same circle are equal) In ∆𝐶𝑃𝐷, ∠𝑃𝐷𝐶 = ∠𝑃𝐶𝐷 =

1 2 180 − 𝑏 = 90 − 𝑏 2 since the

sum of angles of a triangle is 180°.

B A C 𝑐 𝑏 O

In ∆𝐶𝑃𝐵, ∠𝑃𝐵𝐶 = ∠𝑃𝐶𝐵 =

1 2 180 − 𝑐 = 90 − 𝑐 2 since the

sum of angles of a triangle is 180°. ∠𝐵𝐶𝐷 = ∠𝑃𝐶𝐷 + ∠𝑃𝐶𝐵 = 90 −

𝑏 2 + 90 − 𝑐 2

= 180 −

𝑏 2 − 𝑐 2

Since the angle sum at the center of a circle is 360°, reflex ∠𝐵𝑃𝐷 = 360 − 𝑏 − 𝑐 =

1 2 (180 − 𝑏 2 − 𝑐 2)

But 180 −

𝑏 2 − 𝑐 2 = ∠𝐵𝐶𝐷

Therefore reflex ∠𝐵𝑃𝐷 = 2∠𝐵𝐶𝐷

Example 3 Find the size of ∠𝑂𝑁𝑃. P is the center of the circle. Solution ∠𝑂𝑄𝑃 = 2∠𝑂𝑁𝑃 89° = 2∠𝑂𝑁𝑃 ∠𝑂𝑁𝑃=

89 2 = 44.5° M N O P 89°

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

M N P Q

∠𝑄=∠𝑅 Proof ∠𝑁𝑃𝑂 is a central angle subtended by arc MN. ∠𝑁𝑄𝑂 is an inscribed angle subtended by arc MN. Therefore ∠𝑁𝑄𝑂 =

1 2 ∠𝑁𝑃𝑂

∠𝑁𝑅𝑂 is an inscribed angle subtended by arc MN. Therefore ∠𝑁𝑅𝑂 =

1 2 ∠𝑁𝑃𝑂

Thus Therefore ∠𝑁𝑄𝑂 = ∠𝑁𝑅𝑂

M N P Q O

Inscribed angles subtended by equal chords are equal. If 𝑁𝑂 = 𝐾𝐿 then ∠𝑁𝑅𝑂 = ∠𝐿𝑄𝐾.

M N P Q K J

𝑁𝑂 = 𝐾𝐿 Proof Since 𝑁𝑂 and 𝐾𝐿 are equal chords, then they subtend equal angles at the center therefore ∠𝑁𝑃𝑂 = ∠𝐾𝑃𝐿 ∠𝑁𝑅𝑂 =

1 2 ∠𝑁𝑃𝑂 = 1 2 ∠𝐾𝑃𝐿 (Inscribed angle is half the

central angle) ∠𝐿𝑄𝐾 =

1 2 ∠𝐾𝑃𝐿 (inscribed angle is half the central angle)

∠𝐿𝑄𝐾 = ∠𝑁𝑅𝑂

M N P Q K J O

HOMEWORK Find the size of obtuse ∠𝐾𝑃𝑀 in the figure below.

L J K O 65°

ANSWERS TO HOMEWORK

130°

THE END