Theorem 12-9: The measure of an inscribed angles is half the measure of its intercepted arc. m B= 1 / 2 mAC ( B A C.

Theorem 12-9:

The measure of an inscribed angles is half

the measure of its intercepted

arc.

mB=1/2mAC

(

B

A

C

Example”

Find the measure of A

A

B

C

D

900

1100

600

1000

mA=1/2mBCD

(mA=1/2(900+600)mA=1/2(1500)mA=750

Example”

Find the measure of D

A

B

C

D

900

1100

600

1000

mD=1/2mABC

(mD=1/2(1000+900)mD=1/2(1900)mD=950

Corollaries #1

Two inscribed angles that intercept the same arc are congruent.

mBmCB

C

Corollaries #2

An angle inscribed in a semicircle is a right angle

mB=900

B

Corollaries #3

The opposite angles of a quadrilateral inscribed in a circle are supplementary.

mA+mC=1800

mB+mD =1800B

D

A

C

Example”

Find the measure of a and b.

a

b0 O320A is inscribed in a semi-circle, a is a right

angle

Example”

Find the measure of a and b.

a

b0 O320a=900

The sum of the angles of a triangle is 1800, the other angle is 1800-900-320=580

580

Example”

Find the measure of a and b.

a

b0 O320a=900

580=1/2b580

2 21160 =b

Theorem 12-10:

The measure of an angle formed by a

tangent and a chord is half the measure of the intercepted arc.

mC=1/2mBDC

(

B

D

C

Example:

RS and TU are diameters

of A. RB is tangent to

A at point R. Find

mBRT and mTRS.

B

R

U

S

A1260

T

mBRT

B

R

U

S

A1260

T

mBRT=1/2m RT

)

mRT=mURT-mUR

) ) )

mRT=1800-1260

)

mRT=540

)

mBRT=1/2(540)mBRT=270

mTRS

B

R

U

S

A1260

T

mBRS=mBRT+mTRS

270

900=270+mTRS

630=mTRS