1. Properties of a Chord Circle Geometry Homework: Lesson 6.2/1-12, 18
Quiz Friday Lessons 6.1 – 6.2
Ying Yang Project Due Friday

2. 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. What is a chord?
The diameter of a circle is the longest chord of any circle since it passes through the center.
A diameter satisfies the definition of a chord, however, a chord is not necessarily a diameter.
Therefore, every diameter is a chord, but not every chord is a diameter.

4. Chord Property #1, #2 and #3
A perpendicular line from the center of a chord to the center of a circle:
#1: Makes a 90° angle with the chord
#2: Creates two equal line segments RS and QR
#3: Must pass through the center of the circle O

6. is a diameter of the circle.

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

8. AB  CD
if and only if
EF  EG.

9. Chord Arcs Conjecture  if and only if
In the same circle, or in congruent circles, two minor arcs are congruent if and only if their corresponding chords are congruent. 
if and only if

10. 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.  ,

11. 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.

12. Ex: Using Chord Arcs Conj.
Because AD = DC, and = So, m = m
2x°
2x = x + 40
Substitute
Subtract x from each side.
x = 40

13. Ex: Finding the Center of a Circle
Perpendicular Bisector to a Chord Conjecture 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.

14. Ex: Finding the Center of a Circle con’t
Step 2: Draw the perpendicular bisector of each chord. These are the diameters.

15. Ex: Finding the Center of a Circle con’t
Step 3: The perpendicular bisectors intersect at the circle’s center.

16. Chord Distance to the Center Conjecture
In the same circle, or in congruent circles, two chords are congruent if and only if they are equidistant from the center.
AB  CD if and only if EF  EG.

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

18. Ex: Find CF con’t
Because AB and DE are congruent chords, they are equidistant from the center. So CF  CG. To find CG, first find DG.
CG  DE, so CG bisects DE. Because DE = 8, DG = =4.

19. Ex: Find CF con’t
Then use DG to find CG. DG = 4 and CD = 5, so ∆CGD is a right triangle. So CG = 3. Finally, use CG to find CF. Because CF  CG, CF = CG = 3

21. An angle whose vertex is on the center of the circle and whose sides are radii of the circle
An angle whose vertex is ON the circle and whose sides are chords of the circle

22. Arcs are defined by their endpoints
Naming Arcs
Arcs are defined by their endpoints
Minor Arcs require the 2 endpoints
Major Arcs require the 2 endpoints AND a point the arc passes through
Semicircles also require the 2 endpoints (endpoints of the diameter) AND a point the arc passes through K

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

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

27. What can you tell me about segment AC if you know it is the perpendicular bisectors of segments DB?
It’s the DIAMETER!!! A C B

28. Finding the Center of a Circle
If one chord is a perpendicular bisector of another chord, then the first chord is a diameter.
This 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.

29. Finding the Center of a Circle
Step 2: Draw the perpendicular bisector of each chord.
These are diameters of the circle.

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

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

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

33. 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

34. 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.

35. Congruent chords intercept congruent arcs
Congruent chords intercept congruent arcs

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

37. Let’s get crazy…
Find c. 3 8 c 12

38. Let’s get crazy… Find c. Step 1: What do we need to find?3 8 c 12
Step 1: What do we need to find?
We need a radius to complete this big triangle.

39. Let’s get crazy… Find c. How do we find a radius?3 8 c 12
How do we find a radius?
We can draw multiple radii (radiuses).

40. Let’s get crazy… Find c. How do we find a radius?5 4 3 8 c 12
How do we find a radius?
Now what do we have and what will we do?
We create this triangle,
use the chord property to find on side
& the Pythagorean theorem find the missing side.

41. Let’s get crazy… Find c. !!! Pythagorean Theorem !!! c2=a2+b25 4 3 13 c 12
!!! Pythagorean Theorem !!!
c2=a2+b2
c2=52+122
c2=169
c = 13.