1. In the diagram below, O circumscribes quadrilateral ABCD
Inscribed Angles
Lesson 12-3
Lesson Quiz .
In the diagram below, O circumscribes quadrilateral ABCD
and is inscribed in quadrilateral XYZW.
1. Find the measure of each inscribed angle.
2. Find m  DCZ.
3. Are  XAB and  XBA congruent? Explain.
4. Find the angle measures in quadrilateral XYZW.
m A = 100; m B = 75; m C = 80; m D = 105 45
Yes; each is formed by a tangent and a chord, and they intercept the same arc.
m X = 80; m Y = 70; m Z = 90; m W = 120
5. Does a diagonal of quadrilateral ABCD intersect the center of the circle? Explain how you can tell.
No; the diagonal would be a diameter of O and the inscribed angle would
be a right angle, which was not found in Exercise 1 above. .
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2. Angle Measures and Segment Lengths
Lesson 12-4
Check Skills You’ll Need
(For help, go to Lessons 12-1 and 12-3.)
In the diagram at the right, FE and FD are
tangents to C. Find each arc measure, angle
measure, or length.
1. mDE 2. mAED 3. mEBD
4. m EAD 5. m AEC 6. CE
7. DF 8. CF 9. m EFD .
Check Skills You’ll Need
12-4

3. Angle Measures and Segment Lengths
Lesson 12-4
Check Skills You’ll Need
Solutions
1. mDE = m DCE = 57
2. mAED = m ACD = 180
3. mEBD = 360 – m DCE = 360 – 57 = 303
4. Since radii of the same circle are congruent, ACE is isosceles and m ACE =
180 – 57 = 123. The sum of the angles of a triangle is 180 and the base angles
of an isosceles triangle are congruent, so m EAD = m EAD =
180 – 123 = m EAD = 57 ÷ 2 = 28.5
5. From Exercise 4, m AEC = m EAD = 28.5
6. Since all radii of the same circle are congruent, CE = CD = 4.
12-4

4. Angle Measures and Segment Lengths
Lesson 12-4
Check Skills You’ll Need
Solutions (continued)
7. By Theorem 11-3, two tangents tangent to a circle from a point outside
the circle are congruent, so DF = EF = 2.
8. Draw CF. Since FE is tangent to the circle, FE CE . By def. of , CEF
is a right angle. By def. of right angle, m CEF = 90, so CEF is a right
triangle. From Exercise 6, CE = 4. Also, EF = 2. Use the Pythagorean Theorem:
a2 + b2 = c CE 2 + EF 2 = CF = CF = CF 2
20 = CF CF = =
9. CEFD is a quadrilateral, so the sum of its angles is 360. From Exercise 8,
m ACE = 90. Similarly, m CDF = 90. Also, m DCE = 57. So, m EFD +
m FEC + m DCE + m CDF = m EFD = 360
m EFD = m EFD = 360 – 237 = 123
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5. Angle Measures and Segment Lengths
Lesson 12-4
Notes
A secant is a line that intersects a circle at two points.
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6. Angle Measures and Segment Lengths
Lesson 12-4
Notes
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7. Angle Measures and Segment Lengths
Lesson 12-4
Notes
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8. Angle Measures and Segment Lengths
Lesson 12-4
Notes
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9. Angle Measures and Segment Lengths
Lesson 12-4
Notes
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10. Angle Measures and Segment Lengths
Lesson 12-4
Notes
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11. Angle Measures and Segment Lengths
Lesson 12-4
Additional Examples
Finding Angle Measures
Find the value of the variable. a.
x = (268 – 92) The measure of an angle formed by two
lines that intersect outside a circle is half
the difference of the measures of the
intercepted arcs (Theorem (2)). 1 2
x = 88 Simplify.
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12. Angle Measures and Segment Lengths
Lesson 12-4
Additional Examples
(continued) b.
94 = (x + 112) The measure of an angle formed by
two lines that intersect inside a circle
is half the sum of the measures of the
intercepted arcs (Theorem (1)). 1 2
94 = x + 56 Distributive Property 1 2
38 = x Subtract. 1 2
Quick Check
76 = x Multiply each side by 2.
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13. Angle Measures and Segment Lengths
Lesson 12-4
Additional Examples
Real-World Connection
An advertising agency wants a frontal photo of a “flying saucer” ride at an amusement park. The photographer stands at the vertex of the angle formed by tangents to the “flying saucer.” What is the measure of the arc that will be in the photograph?
In the diagram, the photographer stands at point T.
TX and TY intercept minor arc XY and major arc XAY.
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14. Angle Measures and Segment Lengths
Lesson 12-4
Additional Examples
Let mXY = x.
(continued)
Then mXAY = 360 – x.
m T = (mXAY – mXY) The measure of an angle formed by two
lines that intersect outside a circle is half
the difference of the measures of the
intercepted arcs (Theorem (2)). 1 2
72 = [(360 – x) – x] Substitute. 1 2
72 = (360 – 2x) Simplify. 1 2
72 = 180 – x Distributive Property
x + 72 = Solve for x.
x = 108
Quick Check
A 108° arc will be in the advertising agency’s photo.
12-4

15. Angle Measures and Segment Lengths
Lesson 12-4
Additional Examples
Finding Segment Lengths
Find the value of the variable. a.
5 • x = 3 • 7 Along a line, the product of the lengths
of two segments from a point to a circle
is constant (Theorem (1)).
5x = Solve for x.
x = 4.2 b.
8(y + 8) = 152 Along a line, the product of the lengths
of two segments from a point to a circle is
constant (Theorem (3)).
8y + 64 = 225 Solve for y.
8y = 161
y =
Quick Check
12-4

16. Angle Measures and Segment Lengths
Lesson 12-4
Additional Examples
Real-World Connection
A tram travels from point A to point B along the arc of a circle with a radius of 125 ft. Find the shortest distance from point A to point B.
Quick Check
The perpendicular bisector of the chord AB
contains the center of the circle.
Because the radius is 125 ft, the diameter is 2 • 125 = 250 ft.
The length of the other segment along the diameter is 250 ft – 50 ft, or 200 ft.
x • x = 50 • 200 Along a line, the product of the lengths
of the two segments from a point to a
circle is constant (Theorem (1)).
x2 = 10, Solve for x.
x = 100
The shortest distance from point A to point B is 200 ft.
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17. Angle Measures and Segment Lengths
Lesson 12-4
Lesson Quiz
Use M for Exercises 1 and 2.
1. Find a.
2. Find x. .
Use O for Exercises 3–5.
3. Find a and b.
4. Find x to the nearest tenth.
5. Find the diameter of O. . 82
a = 60; b = 28 24
15.5 . 22
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