1. Prove that two parallel secants intercept equal arcs on a circle.

§ 10.1O

E

DC

BA P

GF

Given: AB CD.

Prove: arc AC = arc BD

Statement Reason

1. OE AB. Construction

3. AB CD Given

4. OQC = 90 Alternate interior angles

6. FG OE Construction

2. OPB = 90 Def perpendicular

9. arc AE = arc BE Part 1 theorem done in class.

Q

5. OE CD Definition

7. FG AB CD Line to same line.

8. arc CE = arc DE Part 1 theorem done in class.

10. arc AC = arc BD (8) – (9)

C

BA

O

2. Prove that an inscribed angle, with the center of the circle on the exterior of the angle, is half the measure of the intercepted arc.

Given: Inscribed angle ABC with O on the exterior of the angle.

Prove: ABC = ½ arc AC

Statement Reason

1. Draw diameter BOD Construction

4. ABC = ½ (arc AD - ½ arc DC) = ½ arc AC

Addition

2. ABD = ½ arc AD Case I done in class.

D

3. DBC = ½ arc DC Case I done in class.

O

2. Prove that an inscribed angle, with the center of the circle on the interior of the angle, is half the measure of the intercepted arc.

C

B

A

OGiven: Inscribed angle ABC with O on the

interior of the angle.

Prove: ABC = ½ arc AC

Statement Reason

1. Draw diameter BOD Construction

4. ABC = ½ (arc AD + ½ arc DC) = ½ arc AC

Subtraction

2. ABD = ½ arc AD Case I done in class.

D

3. DBC = ½ arc DC Case I done in class.

3. Prove that an inscribed angle, with the center of the circle on the exterior of the angle, is half the measure of the intercepted arc. KN = 6 tangents from the same point (K) are equal.KO = 6 + 4 = 10JP = 6.LN = 3 a radiusLO = 5 from Pythagorean theorem on triangle LNO a 3-4-5

trianglePO = LO – LP = 5 – 3 = 2KL = √45 from Pythagorean theorem on triangle LNK a 3-6-x

triangle

4

3

6

PM

O

J

L

NK

4. Prove that the angle formed by a secant and a tangent is half the difference of the intercepted arcs.

Given: Secant AB and tangent BC

Prove: B = ½ arc (AC – CD).

Statement Reason

1. Draw AC Construction

3. B = ACE - A Subtraction.

4. ACE = ½ arc AC Angle formed by chord and tangent.

6. B = ½ (arc AC - arc CD). Substitution

2. ACE = A + B. Subtraction.

A

ED

C

B

5. A = ½ arc CD. Inscribed angle

5. Prove that the angle formed by two tangents is half the difference of the intercepted arcs.

Given: Tangent AB and tangent BC

Prove: B = ½ arc (ADC – AC).

Statement Reason

1. Draw AC Construction

3. B = CAE - ACB Subtraction.

4. CAE = ½ arc ADC Angle formed by chord and tangent.

6. B = ½ (arc ADC - arc AC). Substitution

2. CAE = ACB + B. Exterior angle.

A

ED

CB

5. ACB = ½ arc AC. Angle formed by chord and tangent.

6. The sum of the lengths of two tangent segments to a circle form the same exterior point is equal to the diameter of the circle. Find the measure of the angle determined by the tangent segments

DC = CE and DC + CE = AB making each of the tangent segments a length equal to the radius of the circle. This means that ODCE would be a rhombus but since the radii are perpendicular to the tangent segments it is a square and hence the angle at C is 90.

A

E

D C

B

7. If point P is 13 inches from the center of a circle with a 10 inch diameter, how long are the tangent segments from P?

APO

C

B

13

5

5

5

Use the Pythagorean Theorem on triangle OBP with leg of 5 and hypotenuse of 13 to find the length of the other leg or tangent segment.

2 2 213 5 BP and BP 12

8. Two chords of a circle intersect. The segments of one chord have lengths 4 and 6. if the length of one segment of the other chord is 3, find the length of the remaining segment.

Use the two-chord power theorem.

3x = (4)(6) and x = 8.

A 6

D

C

B

4

3

X

9. Solve for x.

Use the theorem that x is half the sum of the intercepted arcs.

x

106

174

1 1x (106 174) (280) 140

2 2

10. Solve for x.

Use the theorem that x is half the difference of the intercepted arcs.

1 1x (200 56) (144) 72

2 2

x

200

F

E

D

G 56

11. Solve for x.

Use the theorem that x is half the difference of the intercepted arcs and the fact that the missing measurement is 360 – 92 or 268.

1 1x (268 92) (176) 88

2 2

92

X

12. Find QR QS when:  

a. QS = 9 and QR = 5 b. QU = 14 and QT = 10c. QT = 1 and TU = 13d. QR = 2 and SR = 7

Use the two-chord power theorem: QR QS = QU QT

a.QR QS = 9 5 = 45.

b.QR QS = QU QT = 14 10 = 140

c.QR QS = QU QT = 12 1 = 12

d.QR QS = 2 5 = 10

13. Segment DB is a diameter of a circle. A tangent through D and a secant through B intersect at a point A. The secant also intersects the circle at C. Prove the DB 2 = (AB)(BC).

AO

C

B

D

From the tangent-secant theorem

we know that

AB AC = AD 2

AB (AB – BC) = AD 2 Since AC = AB – BC

AB 2 – AB BC = AD 2 Distributive property of multiplication.

AB 2 – AD 2 = AB BC Rearrangement of equation.

BD 2 = AB BC Since BD 2 = AB 2 – AD 2 by the Pythagorean Theorem.