OF A CIRCLE 2 I NTRODUCTION Our journey with the geometry of angles - - PowerPoint PPT Presentation

DAY 123 – ANGLE PROPERTIES

OF A CIRCLE2

INTRODUCTION

Our journey with the geometry of angles in a circle

• continues. I have looked at central angles and how

they relate to angles inscribed angles. We would like to go a little further and look at how central angles, inscribed angles and their related with the angles formed due to the intersection of secants (circumscribed angles). In this lesson, we are going to discuss the relationship between central, inscribed, and circumscribed angles.

VOCABULARY

Central Angles Angles whose vertex is the center of the circle and bounded by two radii Inscribed Angles Angles formed at the circumference of the circle when two-chord intersect Circumscribed Angles Angles formed outside the circle when two secants intersect.

Intersecting tangents to one circle We first show that the tangents YX and WX are equal. We will use the about tangents that would be proved later in the following presentations; that tangents intersect at right angles. Consider the circle below.

W X Y Z

Using the property above that tangents and radius meet at right angle, we have triangle ZXY and ZXW being right angles. ZX is common to both and it is the hypotenuse 𝑎𝑍 = 𝑎𝑋 radius to the circle Hence the Hypotenuse –leg postulate for congruence triangle is satisfied implying that right the two triangles are congruent. Therefore, Corresponding sides and angles are equal. This means that the tangents are equal, 𝑍𝑌 = 𝑋𝑌

We would like also to come up with more results. Since corresponding angles are equal, we have that ∠𝑍𝑌𝑎 = ∠𝑋𝑌𝑎 ∠𝑍𝑎𝑌 = ∠𝑋𝑎𝑌 This implies that the line ZX is a bisector

• f ∠𝑍𝑎W and ∠𝑋𝑌𝑍

Also, since angles ZWX and ZYX are all right angles, the remaining two angles in quadrilateral WZYX add up to 360° − 90° − 90° = 180°

Therefore, the result is that when two tangents to a circle intersect at a common point, the angles formed at their intersection and at the intersection of the two radii drawn from the point where they touch the circle are supplementary and are bisected by the line connecting the center to the intersection of the tangents.

We would also want to make a proof that we will use in this lesson that; Angle at the intersection of the tangent and the chord is equal to the inscribed angle in the opposite segment subtended by the same chord Consider the figure below where WX is the tangent to the circle at W and ∠𝑁𝑋𝑌 = 𝛽

𝛽 W X O T M

Since OW is the radius, ∠𝑃𝑋𝑌 = 90° since the tangent is always perpendicular to the radius. Thus, ∠𝑃𝑋𝑁 = 90° − 𝛽 Since 𝑃𝑋 = 𝑃𝑁(𝑠𝑏𝑒𝑗𝑣𝑡); OWM is Iscosceles triangle with base angles at W and M, thus ∠𝑃𝑁𝑋 = 90° − 𝛽 ∠𝑋𝑃𝑁 = 180° − 90° − 𝛽 − 90° − 𝛽 = 2𝛽 ∠𝑋𝑈𝑁 = 1

2 ∠𝑋𝑃𝑁 = 𝛽 (central and inscribed angle

subtended by the common chord on one side of the chord) Thus ∠𝑋𝑈𝑁 = ∠𝑌𝑋𝑁 = 𝛽 as required Something important noting here is that the central angle subtended by a chord is twice the angle the chord makes with the tangent.

Angles as a result of intersection of a secant and a tangent

1 2 ∠𝐵𝑃𝑁 = ∠𝐶𝐵𝑁 (Central angle to a chord and angle at

intersection of the chord and the tangent) ∠𝑁𝐶𝐵 = ∠𝐷𝑁𝐵 − ∠𝐶𝐵𝑁 (Interior and exterior angles of a triangle) ∠𝐷𝑁𝐵 = 1

2 ∠𝐷𝑃𝐵

(Inscribed and centrall angle from the common chord)

A B C M O

Upon substitution, we get that ∠𝑵𝑪𝑩 = 𝟐 𝟑 ∠𝑫𝑷𝑩 − 𝟐 𝟑 ∠𝑩𝑷𝑵 = 𝟐 𝟑 (∠𝑫𝑷𝑩 − ∠𝑩𝑷𝑵) Angles as a result of intersection of two secants Consider the diagram above

D A B C M O

∠𝑁𝐶𝐵 = ∠𝐸𝐵𝐷 − ∠𝐵𝐷𝑁(Interior and exterior angles

• f a triangle)

But ∠𝐸𝐵𝐷 =

1 2 ∠𝐸𝑃𝐷 (Central and inscribed angle of

a common chord) Also, ∠𝐵𝐷𝑁 =

1 2 ∠𝐵𝑃𝑁 (Central and inscribed angle

• f a common chord)

Upon substitution, we get ∠𝑵𝑪𝑩 = 𝟐 𝟑 ∠𝑬𝑷𝑫 − 𝟐 𝟑 ∠𝑩𝑷𝑵 = 𝟐 𝟑 (∠𝑬𝑷𝑫 − ∠𝑩𝑷𝑵)

Example In the figure below, O is the center of the circle. Using the angles given, find the measure of angle P. Solution ∠𝑍𝑈𝑃 = ∠𝑈𝑍𝑇 = 10° (Base angle of Isosceles triangle since 𝑈𝑃 = 𝑃𝑍(radius)) ∠𝑈𝑃𝑍 = 180° − 10° − 10° = 160° (Interior angles of a triangle)

S Y P T F O 43°

10°

∠𝑍𝑃𝐺 = 2∠𝑄𝑍𝐺 = 2 × 43° = 86° (Central angle to a chord and angle at the intersection of a chord and the tangent) Using properties of angles when tangent and secant intersects, we have ∠𝑄 = 1 2 ∠𝑈𝑃𝑍 = ∠𝑍𝑃𝐺 = 1 2 160 − 86 = 37°

HOMEWORK Given that O is the center of the circle given, find the size of angle NGO.

A H O 38° 130° G N

ANSWERS TO HOMEWORK

63°

THE END