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 𝑍𝑌=𝑍𝑊 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, 𝑌𝑋=𝑊𝑋

We would like also to come up with more results We would like also to come up with more results. Since corresponding angles are equal, we have that ∠𝑌𝑋𝑍=∠𝑊𝑋𝑍 ∠𝑌𝑍𝑋=∠𝑊𝑍𝑋 This implies that the line ZX is a bisector of ∠𝑌𝑍W and ∠𝑊𝑋𝑌 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 ∠𝑀𝑊𝑋=𝛼 𝛼 W X O T M

Since OW is the radius, ∠𝑂𝑊𝑋= 90° since the tangent is always perpendicular to the radius. Thus, ∠𝑂𝑊𝑀=90°−𝛼 Since 𝑂𝑊=𝑂𝑀(𝑟𝑎𝑑𝑖𝑢𝑠); OWM is Iscosceles triangle with base angles at W and M, thus ∠𝑂𝑀𝑊=90°−𝛼 ∠𝑊𝑂𝑀=180°− 90°−𝛼 − 90°−𝛼 =2𝛼 ∠𝑊𝑇𝑀= 1 2 ∠𝑊𝑂𝑀=𝛼 (central and inscribed angle subtended by the common chord on one side of the chord) Thus ∠𝑊𝑇𝑀=∠𝑋𝑊𝑀=𝛼 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 ∠𝐴𝑂𝑀=∠𝐵𝐴𝑀 (Central angle to a chord and angle at intersection of the chord and the tangent) ∠𝑀𝐵𝐴=∠𝐶𝑀𝐴−∠𝐵𝐴𝑀 (Interior and exterior angles of a triangle) ∠𝐶𝑀𝐴= 1 2 ∠𝐶𝑂𝐴 (Inscribed and centrall angle from the common chord) A B C M O

Upon substitution, we get that ∠𝑴𝑩𝑨= 𝟏 𝟐 ∠𝑪𝑶𝑨− 𝟏 𝟐 ∠𝑨𝑶𝑴= 𝟏 𝟐 (∠𝑪𝑶𝑨−∠𝑨𝑶𝑴) Angles as a result of intersection of two secants Consider the diagram above D A B C M O

∠𝑀𝐵𝐴=∠𝐷𝐴𝐶−∠𝐴𝐶𝑀(Interior and exterior angles of a triangle) But ∠𝐷𝐴𝐶= 1 2 ∠𝐷𝑂𝐶 (Central and inscribed angle of a common chord) Also, ∠𝐴𝐶𝑀= 1 2 ∠𝐴𝑂𝑀 (Central and inscribed angle of a common chord) Upon substitution, we get ∠𝑴𝑩𝑨= 𝟏 𝟐 ∠𝑫𝑶𝑪− 𝟏 𝟐 ∠𝑨𝑶𝑴= 𝟏 𝟐 (∠𝑫𝑶𝑪−∠𝑨𝑶𝑴)

Example In the figure below, O is the center of the circle Example In the figure below, O is the center of the circle. Using the angles given, find the measure of angle P. Solution ∠𝑌𝑇𝑂=∠𝑇𝑌𝑆=10° (Base angle of Isosceles triangle since 𝑇𝑂=𝑂𝑌(radius)) ∠𝑇𝑂𝑌=180°−10°−10°=160° (Interior angles of a triangle) S Y P T F O 43° 10°

∠𝑌𝑂𝐹=2∠𝑃𝑌𝐹=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 ∠𝑃= 1 2 ∠𝑇𝑂𝑌=∠𝑌𝑂𝐹 = 1 2 160−86 =37°

homework Given that O is the center of the circle given, find the size of angle NGO. K A H O 38° 130° G N

Answers to homework 63°

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