1. Lesson 6.2 Properties of Chords
Objective: Discover properties of chords of a circle
Homework: Lesson 6.2/1-12

2. Any chord divides the circle into two arcs.
What is a chord?
A chord is a segment with endpoints on a circle.
Any chord divides the circle into two arcs.
A diameter divides a circle into two semicircles.
Any other chord divides a circle into a minor arc and a major arc.

3. Chord Arcs Conjecture
In the same circle, two minor arcs are congruent if and only if their corresponding chords are congruent.
IFF G
and
IFF
and

5. Perpendicular Bisector of a Chord Conjecture
If a diameter of a circle is perpendicular to a chord, then the diameter bisects the chord and its arc. H

6. is a diameter of the circle.

7. Perpendicular Bisector to a Chord Conjecture
If one chord is a perpendicular bisector of another chord,
then the first chord passes through the center of the circle and is a diameter.
is a diameter of the circle.

8. If one chord is a perpendicular bisector of another chord, then the first chord is a diameter. ,

9. Ex. 4: Using Chord Arcs Conjecture
2x° C A
2x = x + 40 B
x = 40

10. Ex. 5: Finding the Center of a Circle
Perpendicular bisector to a chord can be used to locate a circle’s center as shown in the next few slides.
Step 1: Draw any two chords that are not parallel to each other.

11. Ex. 5: Finding the Center of a Circle
Step 2: Draw the perpendicular bisector of each chord.
These are the diameters.

12. Ex. 5: Finding the Center of a Circle
Step 3: The perpendicular bisectors intersect at the circle’s center.

13. Chord Distance to the Center Conjecture

14. AB  CD
if and only if
EF  EG.

15. Ex. 7:
AB = 8; DE = 8, and CD = 5. Find CF.

18. (x + 40)° D
2x°
2x = x + 40
x = 40

19. Ex.4: Solve for the missing sides.A 7m 3m C D B
BC =
AB =
AD ≈ 7m
14m
7.6m

21. Ex.6: QR = ST = 16. Find CU.
x = 3

22. Ex 7: AB = 8; DE = 8, and CD = 5. Find CF.
CG = CF
CG = 3 = CF

23. Diameter is the perpendicular bisector of the chord
Ex.8: Find the length of Tell what theorem you used.
BF = 10
Diameter is the perpendicular bisector of the chord
Therefore, DF = BF

24. Congruent chords are equidistant from the center.
Ex.9: PV = PW, QR = 2x + 6, and ST = 3x – 1. Find QR.
Congruent chords are equidistant from the center.

25. Congruent chords intercept congruent arcs
Congruent chords intercept congruent arcs

26. Congruent chords are equidistant from the center.
Ex.11:
Congruent chords are equidistant from the center.