Post on 29-Dec-2015
Chapter 10 – Circles
Section 10.3 – Inscribed Angles
Unit Goal
Use inscribed angles to solve problems.
AB
C
D
Basic Definitions
INSCRIBED ANGLE – an angle whose vertex is on the circleINTERCEPTED ARC – the arc whose endpoints are are on the inscribed angle
AB
C
D
is an inscribed angle.DCB
is the intercepted arc.DB
What Is the Measure ofan Inscribed Circle?
What is the measure of ?What is the measure of ?
BECBDC
A
B
C
D
E
Theorem 10.8Measure of an Inscribed Angle
The measure of an inscribed angle is ½ of its intercepted arc.
A
B
DC
1
2m BDC mBC
Example
Find the measure of the angle or arc:
140m BDC
mBIC
A
B
D
C
I
m CDB
20º
A
B
D
C
Example
Find the measure of the angle or arc:
m BDC
mBEC
A
B
D
C
E
AB
D
C
mBC
50º
Example
B
D C
J
K
60º
m BDC m BJC m BKC
Theorem 10.9
If two inscribed angles of a circle intercept the same arc, then the angles are congruent.
B
D C
J
K
60º
Properties of Inscribed Polygons
If all the vertices of a polygon lie on a circle, the polygon is INSCRIBED in the circles and the circle is CIRCUMSCRIBED about the polygon
Theorems About Inscribed Polygons
Theorem 10.10If a right triangle is inscribed in a circle, then the hypotenuse is a diameter of the circle. Conversely, if one side of an inscribed triangle is a diameter of the circle, then the triangle is a right triangle and the angle opposite the diameter is the right angle<B is a right angle iff segment AC is a diameter of the circle C
P
A
B
Theorem 10.11
A quadrilateral can be inscribed in a circle iff its opposite angles are supplementaryD, E, F, and G lie on some circle C iff
m<D + m<F = 180° ANDm<E + m<G = 180°
C
E F
GD
Example
In the diagram, ABCD is inscribed in circle P. Find the measure of each angle.ABCD is inscribed in a circle, so opposite angles are supplementary3x + 3y = 180 and 5x+ 2y = 1803x + 3y = 180 (solve for x) - 3y -3y3x = -3y + 1803 3x = -y + 60Substitute
5x
3x3y
2y
P
A
B
C
D
Substitute this into the second equation 5x + 2y = 180
5 (-y + 60) + 2y = 180
-5y + 300 + 2y = 180
-3y = -120
y = 40x = -y + 60
x = -40 + 60 = 20
Example (cont.)
x = 20, y = 40m<A = 2y, m<B = 3x, m<C = 5x, m<D = 3ym<A = 80°m<B = 60°m<C = 100°m<D = 120°
HW Assignment
p. 616-617 (4 – 28 even)