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)