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Inscribed AnglesInscribed Angles
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Given:  𝐺𝐻  and the tangent l             intersecting at 𝐻 on circle ⨀𝐸
Prove: 𝑚∠𝐺𝐻𝐼= 1 2  𝑚 𝐺𝐹𝐻
Challenge ProblemChallenge Problem
F
G
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l
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 𝐺𝐻  and tangent line l intersect at 𝐻 on ⨀ 𝐸 (given)
Construct diameter  𝐻𝐷  intersecting ⨀𝐸 at 𝐷.
∠𝐷𝐻𝐼 is a right angle (tangent and radius are perpendicular)
 𝐷𝐺𝐻  is a semicircle of measure 180 (def of semicircle)
𝑚∠𝐷𝐻𝐺 +𝑚∠𝐺𝐻𝐼 =𝑚∠𝐷𝐻𝐼 (angle addition)
𝑚 𝐷𝐺 +𝑚 𝐺𝐹𝐻 =𝑚 𝐷𝐺𝐻  (arc addition)
90=𝑚∠𝐷𝐻𝐺+𝑚∠𝐺𝐻𝐼 (substitution)
180=𝑚 𝐷𝐺 +𝑚 𝐺𝐹𝐻  (substitution)
90= 1 2  (𝑚 𝐷𝐺  + 𝑚 𝐺𝐹𝐻 ) (division)
𝑚∠𝐷𝐻𝐺 + 𝑚∠𝐺𝐻𝐼 =  1 2  𝑚 𝐷𝐺  +  1 2 𝑚 𝐺𝐹𝐻  (substitution and distribution)
𝑚∠𝐷𝐻𝐺 =  1 2  𝑚 𝐷𝐺  (Inscribed angle theorem)
𝑚∠𝐺𝐻𝐼 =  1 2  𝑚 𝐺𝐹𝐻  (subtraction)
D
F
G
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l
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ObjectivesObjectives
Use inscribed angles to solve problems.
Use properties of inscribed polygons.
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Using Inscribed AnglesUsing Inscribed Angles
An inscribed angle is anangle whose vertex is ona circle and whose sidescontain chords of thecircle.  The arc that lies inthe interior of an inscribedangle and has endpointson the angle is called theintercepted arc of theangle.
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Measure of an Inscribed AngleMeasure of an Inscribed Angle
If an angle is inscribed in a circle, then its measure is one half the measure of its intercepted arc.
𝑚𝐴𝐷𝐵 = ½𝑚
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Finding Measures of Arcs and Inscribed AnglesFinding Measures of Arcs and Inscribed Angles
Find the measureof the blue arc orangle.
m          = 2mQRS =
2(90°) = 180°
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Finding Measures of Arcs and Inscribed AnglesFinding Measures of Arcs and Inscribed Angles
Find the measureof the blue arc orangle.
m          = 2mZYX =
2(115°) = 230°
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Ex. 1:  Finding Measures of Arcs and Inscribed AnglesEx. 1:  Finding Measures of Arcs and Inscribed Angles
Find the measureof the blue arc orangle.
m          = ½ m
½ (100°) = 50°
100°
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Ex. 2:  Comparing Measures of Inscribed AnglesEx. 2:  Comparing Measures of Inscribed Angles
Find mACB,mADB, andmAEB.
The measure of eachangle is half themeasure of
m       = 60°, so themeasure of eachangle is 30°
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TheoremTheorem
If two inscribed anglesof a circle interceptthe same arc, then theangles are congruent.
 D
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Ex. 3:  Finding the Measure of an AngleEx. 3:  Finding the Measure of an Angle
It is given thatmE = 75°.  Whatis mF?
E and F bothintercept         , so F.  So,mF = mE = 75°
75°
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Using Properties of Inscribed PolygonsUsing Properties of Inscribed Polygons
If all of the vertices ofa polygon lie on acircle, the polygon isinscribed in the circleand the circle iscircumscribed aboutthe polygon. Thepolygon is aninscribed polygonand the circle is acircumscribed circle.
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TheoremTheorem
If a right triangle is inscribed in a circle,then the hypotenuse is a diameter ofthe circle.  Conversely, if one side ofan inscribed triangle is a diameter ofthe circle, then the triangle is a righttriangle and the angle opposite thediameter is the right angle.
B is a right angle if and only if AC isa diameter of the circle.
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TheoremTheorem
A quadrilateral can be inscribed in a circle if and only if its opposite angles are supplementary.
D, E, F, and G lie on some circle, C, if and only if 𝑚𝐷 + 𝑚𝐹 = 180° 
𝑎𝑛𝑑 𝑚𝐸 + 𝑚𝐺 = 180°
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Ex. 5:  Using TheoremsEx. 5:  Using Theorems
Find the value of each variable.  

AB is a diameter.  So, C is a right angle and mC = 90°
2𝑥° = 90°
𝑥 = 45
2x°
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Ex. 5:  Using TheoremsEx. 5:  Using Theorems
Find the value of each variable.  

DEFG is inscribed in a circle, so opposite angles are supplementary.
𝑚𝐷 + 𝑚𝐹 = 180°
𝑧 + 80 = 180
𝑧 = 100
120°
80°
y°
z°
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Ex. 5:  Using TheoremsEx. 5:  Using Theorems
Find the value of each variable.  

DEFG is inscribed in a circle, so opposite angles are supplementary.
𝑚𝐸 + 𝑚𝐺 = 180°
𝑦 + 120 = 180
𝑦 = 60
120°
80°
y°
z°
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Ex. 6:  Using an Inscribed QuadrilateralEx. 6:  Using an Inscribed Quadrilateral
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Ex. 6:  Using an Inscribed QuadrilateralEx. 6:  Using an Inscribed Quadrilateral
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Angle Measures and Segment Lengthsin CirclesAngle Measures and Segment Lengthsin Circles
Objectives:
1.To find the measures of s formed by chords,secants, & tangents.
2.To find the lengths of segments associated withcircles.
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SecantsSecants
E
A
B
F
Secant – A line that intersects a circle in exactly 2 points.
 𝐸𝐹  or  𝐴𝐵  are secants
 𝐴𝐵  is a chord
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Theorem.  The measure of an  formed by 2 lines thatintersect inside a circle is
𝑚1 =  1 2 (𝑥 + 𝑦)
Measure of intercepted arcs
𝟏
𝒙°
𝒚°
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Theorem.  The measure of an  formed by 2 linesthat intersect outside a circle is
𝒎𝟏 =  𝟏 𝟐 (𝒙 − 𝒚)
Smaller Arc
Larger Arc
x°
y°
1
x°
y°
1
2 Secants:
x°
y°
1
Tangent & a Secant
2 Tangents
3 cases:
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Ex.1 2:Ex.1 2:
Find the measure of arc 𝑥.
Find the 𝑚𝑥.
𝑚1 =  1 2 (𝑥 + 𝑦)
94 =  1 2 (112 + 𝑥)
188 = (112 + 𝑥)
76° = 𝑥
94°
112°
𝒙°
68°
104°
92°
268°
𝒙°
𝑚𝑥 = ½(𝑥 − 𝑦)
𝑚𝑥 = ½(268 − 92)
𝑚𝑥 = ½(176)
𝑚𝑥 = 88°
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Lengths of Secants, Tangents, ChordsLengths of Secants, Tangents, Chords
2 Chords
𝑎•𝑏 = 𝑐•𝑑
2 Secants
𝑎
𝑐
𝑏
𝑑
𝑥
𝑤
𝑧
𝑦
𝑤(𝑤 + 𝑥) = 𝑦(𝑦 + 𝑧)
Tangent & Secant
𝑡
𝑦
𝑧
𝑡2 = 𝑦(𝑦 + 𝑧)
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Ex. 4Ex. 4
Find length of 𝑥.
Find the length of 𝑔.
3
𝑥
7
5
𝑎•𝑏 = 𝑐•𝑑
(3)•(7) = (𝑥)•(5)
21 = 5𝑥
4.2 = 𝑥
15
8
𝑔
𝑡2 = 𝑦(𝑦 + 𝑧)
152 = 8(8 + 𝑔)
225 = 64 + 8𝑔
161 = 8𝑔
20.125 = 𝑔
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Ex.5:  SecantsEx.5:  Secants
Find the length of 𝑥.
14
20
16
𝑥
𝑤(𝑤 + 𝑥) = 𝑦(𝑦 + 𝑧)
14(14 + 20) = 16(16 + 𝑥)
(34)(14) = 256 + 16𝑥
476 = 256 + 16𝑥
220 = 16𝑥
13.75 = 𝑥
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Ex.6:  little bit of everything!Ex.6:  little bit of everything!
𝑟
9
12
𝑘
8
𝑎°
60°
175°
Solve for 𝒌 first.
𝑤(𝑤 + 𝑥) = 𝑦(𝑦 + 𝑧)
9(9 + 12) = 8(8 + 𝑘)
186 = 64 + 8𝑘
𝑘 = 15.6
Next solve for 𝒓
𝑡2 = 𝑦(𝑦 + 𝑧)
𝑟2 = 8(8 + 15.6)
𝑟2 = 189
𝑟 = 13.7
Lastly solve for 𝒎𝒂
𝑚1 = ½(𝑥 − 𝑦)
𝑚𝑎 = ½(175 – 60)
𝑚𝑎 = 57.5°
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What have we learned??What have we learned??
When dealing with angle measures formed byintersecting secants or tangents you either addor subtract the intercepted arcs depending onwhere the lines intersect.
There are 3 formulas to solve for segmentslengths inside of circles, it depends on whichsegments you are dealing with:  Secants,Chords, or Tangents.
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Q
G
F
D
E
C
1
2
3
4
5
6
A
100°
30°
25°
 𝐴𝐹  is a diameter,
𝑚 𝐴𝐺 =100°,
𝑚 𝐶𝐸 =30°
𝑚 𝐸𝐹 =25°

𝐹𝑖𝑛𝑑 𝑡ℎ𝑒 𝑚𝑒𝑎𝑠𝑢𝑟𝑒 𝑜𝑓 𝑎𝑙𝑙
 𝑛𝑢𝑚𝑏𝑒𝑟𝑒𝑑 𝑎𝑛𝑔𝑙𝑒𝑠
Challenge #1Challenge #1
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Find the value of the variable.
6
𝑦
7
8
8
16
𝑧
Challenge #2Challenge #2
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Find the value of the variable.
16
20
14
𝑥
7
𝑚
3
6.5
Challenge #3Challenge #3
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Q
G
F
D
E
C
1
2
3
4
5
6
A
100°
30°
25°
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Find the value of the variable.
6
𝑦
7
8
8
16
𝑧
(6 + 8) 6 = (7 + 𝑦) 7
84 = 49 + 7𝑦
35 = 7𝑦
𝑦 = 5
𝑧2 = (16 + 8) 8
𝑧2 = 192
𝑧 = 13.9
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Find the value of the variable to the nearest tenth.
16
20
14
𝑥
(20 + 14) 14 = (𝑥 + 16) 16
476 = 16𝑥 + 256
220 = 16𝑥
𝑥 = 13.8
7
𝑚
3
6.5
6.5𝑚 = 3 ∙ 7
6.5𝑚 = 21
𝑚 = 3.2