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

D AY 122 – A NGLE PROPERTIES OF A CIRCLE 1

I NTRODUCTION 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.

V OCABULARY 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. C A D B The line segments AB and CD are chords of the circle.

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

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

Proof We want to prove that given that OM ⊥ 𝐾𝑀 then, 𝐾𝐿 = M 𝐿𝑀. J L K O 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.

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. A 𝑦 O C D 𝑦 B Since the two chords are equidistance from the center of the circle, then 𝐵𝐸 = 𝐶𝐷.

Proof We want to prove that 𝐵𝐸 = 𝐷𝐶 if 𝑃𝑄 = 𝑃𝑁. A 𝑁 C D O 𝑄 B 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.

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 P Q N O M 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

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

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. B C A AB and BC are two chords which meet at point B making an angle. ∠𝐶 is an inscribed angle.

The central angle is twice the size of the inscribed angle subtended by the same arc. B C 𝛽 𝛽 𝜄 O 𝜄 A ∠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.

∠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. C 𝑏 O 𝑐 B A 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) 1 𝑏 In ∆𝐶𝑃𝐷, ∠𝑃𝐷𝐶 = ∠𝑃𝐶𝐷 = 2 since the 2 180 − 𝑏 = 90 − sum of angles of a triangle is 180 ° .

1 𝑐 In ∆𝐶𝑃𝐵, ∠𝑃𝐵𝐶 = ∠𝑃𝐶𝐵 = 2 since the 2 180 − 𝑐 = 90 − sum of angles of a triangle is 180 ° . ∠𝐵𝐶𝐷 = ∠𝑃𝐶𝐷 + ∠𝑃𝐶𝐵 𝑏 𝑐 = 90 − 2 + 90 − 2 𝑏 𝑐 = 180 − 2 − 2 Since the angle sum at the center of a circle is 360 ° , reflex ∠𝐵𝑃𝐷 = 360 − 𝑏 − 𝑐 1 𝑏 𝑐 = 2 (180 − 2 − 2 ) 𝑏 𝑐 But 180 − 2 = ∠𝐵𝐶𝐷 2 − Therefore reflex ∠𝐵𝑃𝐷 = 2∠𝐵𝐶𝐷

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

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

∠𝑄 = ∠𝑅 P Q Proof O M N ∠𝑁𝑃𝑂 is a central angle subtended by arc MN. ∠𝑁𝑄𝑂 is an inscribed angle subtended by arc MN. 1 Therefore ∠𝑁𝑄𝑂 = 2 ∠𝑁𝑃𝑂 ∠𝑁𝑅𝑂 is an inscribed angle subtended by arc MN. 1 Therefore ∠𝑁𝑅𝑂 = 2 ∠𝑁𝑃𝑂 Thus Therefore ∠𝑁𝑄𝑂 = ∠𝑁𝑅𝑂

Inscribed angles subtended by equal chords are equal. P Q N K M J If 𝑁𝑂 = 𝐾𝐿 then ∠𝑁𝑅𝑂 = ∠𝐿𝑄𝐾.

𝑁𝑂 = 𝐾𝐿 P Proof Q O N K M J Since 𝑁𝑂 and 𝐾𝐿 are equal chords, then they subtend equal angles at the center therefore ∠𝑁𝑃𝑂 = ∠𝐾𝑃𝐿 1 1 2 ∠𝐾𝑃𝐿 (Inscribed angle is half the ∠𝑁𝑅𝑂 = 2 ∠𝑁𝑃𝑂 = central angle) 1 2 ∠𝐾𝑃𝐿 (inscribed angle is half the central angle) ∠𝐿𝑄𝐾 = ∠𝐿𝑄𝐾 = ∠𝑁𝑅𝑂

HOMEWORK Find the size of obtuse ∠𝐾𝑃𝑀 in the figure below. K L 65 ° O J

A NSWERS TO HOMEWORK 130 °

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