1. 5-1 Perpendicular and Angle Bisectors Warm Up Lesson Presentation
Lesson Quiz
Holt Geometry

2. Warm Up Construct each of the following. 1. A perpendicular bisector.
2. An angle bisector.
3. Find the midpoint and slope of the segment
(2, 8) and (–4, 6).

3. Objectives Prove and apply theorems about perpendicular bisectors.
Prove and apply theorems about angle bisectors.

4. Vocabulary
equidistant
locus

5. When a point is the same distance from two or more
objects, the point is said to be equidistant from
the objects. Triangle congruence theorems can be
used to prove theorems about equidistant points.

7. A locus is a set of points that satisfies a given condition
A locus is a set of points that satisfies a given condition. The perpendicular bisector of a segment can be defined as the locus of points in a plane that are equidistant from the endpoints of the segment.

8. Example 1A: Applying the Perpendicular Bisector Theorem and Its Converse
Find each measure. MN
MN = LN
 Bisector Thm.
MN = 2.6
Substitute 2.6 for LN.

9. Example 1B: Applying the Perpendicular Bisector Theorem and Its Converse
Find each measure. BC
Since AB = AC and , is the perpendicular bisector of by the Converse of the Perpendicular Bisector Theorem.
BC = 2CD
Def. of seg. bisector.
BC = 2(12) = 24
Substitute 12 for CD.

10. Example 1C: Applying the Perpendicular Bisector Theorem and Its Converse
Find each measure. TU
TU = UV
 Bisector Thm.
3x + 9 = 7x – 17
Substitute the given values.
9 = 4x – 17
Subtract 3x from both sides.
26 = 4x
Add 17 to both sides.
6.5 = x
Divide both sides by 4.
So TU = 3(6.5) + 9 = 28.5.

11. Check It Out! Example 1a
Find the measure.
Given that line ℓ is the perpendicular bisector of DE and EG = 14.6, find DG.
DG = EG
 Bisector Thm.
DG = 14.6
Substitute 14.6 for EG.

12. Check It Out! Example 1b
Find the measure.
Given that DE = 20.8, DG = 36.4, and EG =36.4, find EF.
Since DG = EG and , is the perpendicular bisector of by the Converse of the Perpendicular Bisector Theorem.
DE = 2EF
Def. of seg. bisector.
20.8 = 2EF
Substitute 20.8 for DE.
10.4 = EF
Divide both sides by 2.

13. Remember that the distance between a point and a line is the length of the perpendicular segment from the point to the line.

15. Based on these theorems, an angle bisector can be defined as the locus of all points in the interior of the angle that are equidistant from the sides of the angle.

16. Example 2A: Applying the Angle Bisector Theorem
Find the measure. BC
BC = DC
 Bisector Thm.
BC = 7.2
Substitute 7.2 for DC.

17. Example 2B: Applying the Angle Bisector Theorem
Find the measure.
mEFH, given that mEFG = 50°.
Since EH = GH,
and , bisects
EFG by the Converse
of the Angle Bisector Theorem.
Def. of  bisector
Substitute 50° for mEFG.

18. Example 2C: Applying the Angle Bisector Theorem
Find mMKL.
, bisects JKL
Since, JM = LM, and
by the Converse of the Angle
Bisector Theorem.
mMKL = mJKM
Def. of  bisector
3a + 20 = 2a + 26
Substitute the given values.
a + 20 = 26
Subtract 2a from both sides.
a = 6
Subtract 20 from both sides.
So mMKL = [2(6) + 26]° = 38°

19. Check It Out! Example 2a
Given that YW bisects XYZ and
WZ = 3.05, find WX.
WX = WZ
 Bisector Thm.
WX = 3.05
Substitute 3.05 for WZ.
So WX = 3.05

20. Given that mWYZ = 63°, XW = 5.7, and ZW = 5.7, find mXYZ.
Check It Out! Example 2b
Given that mWYZ = 63°, XW = 5.7, and ZW = 5.7, find mXYZ.
mWYZ + mWYX = mXYZ
 Bisector Thm.
mWYZ = mWYX
Substitute m WYZ for mWYX .
mWYZ + mWYZ = mXYZ
2mWYZ = mXYZ
Simplify.
2(63°) = mXYZ
Substitute 63° for mWYZ .
126° = mXYZ
Simplfiy .

21. Example 3: Application
John wants to hang a spotlight along the back of a display case. Wires AD and CD are the same length, and A and C are equidistant from B. How do the wires keep the spotlight centered?
It is given that So D is on the perpendicular bisector of by the Converse of the Angle Bisector Theorem. Since B is the midpoint of , is the perpendicular bisector of Therefore the spotlight remains centered under the mounting.

22. Check It Out! Example 3
S is equidistant from each pair of suspension lines. What can you conclude about QS?
QS bisects  PQR.

23. Example 4: Writing Equations of Bisectors in the Coordinate Plane
Write an equation in point-slope form for the perpendicular bisector of the segment with endpoints C(6, –5) and D(10, 1).
Step 1 Graph .
The perpendicular bisector of is perpendicular to at its midpoint.

24. Example 4 Continued
Step 2 Find the midpoint of .
Midpoint formula.
mdpt. of =

25. Example 4 Continued
Step 3 Find the slope of the perpendicular bisector.
Slope formula.
Since the slopes of perpendicular lines are opposite reciprocals, the slope of the perpendicular bisector is

26. Example 4 Continued
Step 4 Use point-slope form to write an equation. The perpendicular bisector of has slope and passes through (8, –2).
y – y1 = m(x – x1)
Point-slope form
Substitute –2 for y1, for m, and 8 for x1.

27. Example 4 Continued

28. Check It Out! Example 4
Write an equation in point-slope form for the perpendicular bisector of the segment with endpoints P(5, 2) and Q(1, –4).
Step 1 Graph PQ.
The perpendicular bisector of is perpendicular to at its midpoint.

29. Check It Out! Example 4 Continued
Step 2 Find the midpoint of PQ.
Midpoint formula.

30. Check It Out! Example 4 Continued
Step 3 Find the slope of the perpendicular bisector.
Slope formula.
Since the slopes of perpendicular lines are
opposite reciprocals, the slope of the
perpendicular bisector is

31. Check It Out! Example 4 Continued
Step 4 Use point-slope form to write an equation.
The perpendicular bisector of PQ has slope and passes through (3, –1).
y – y1 = m(x – x1)
Point-slope form
Substitute.

32. Objectives
Prove and apply properties of perpendicular bisectors of a triangle.
Prove and apply properties of angle bisectors of a triangle.

33. Vocabulary concurrent point of concurrency circumcenter of a triangle
circumscribed
incenter of a triangle
inscribed

34. Since a triangle has three sides, it has three perpendicular bisectors
Since a triangle has three sides, it has three perpendicular bisectors. When you construct the perpendicular bisectors, you find that they have an interesting property.

35. The perpendicular bisector of a side of a triangle does not always pass through the opposite vertex.
Helpful Hint

36. When three or more lines intersect at one point, the lines are said to be concurrent. The point of concurrency is the point where they intersect. In the construction, you saw that the three perpendicular bisectors of a triangle are concurrent. This point of concurrency is the circumcenter of the triangle.

37. The circumcenter can be inside the triangle, outside the triangle, or on the triangle.

38. The circumcenter of ΔABC is the center of its circumscribed circle
The circumcenter of ΔABC is the center of its circumscribed circle. A circle that contains all the vertices of a polygon is circumscribed about the polygon.

39. Example 1: Using Properties of Perpendicular Bisectors
DG, EG, and FG are the perpendicular bisectors of ∆ABC. Find GC.
G is the circumcenter of ∆ABC. By the Circumcenter Theorem, G is equidistant from the vertices of
∆ABC.
GC = CB
Circumcenter Thm.
GC = 13.4
Substitute 13.4 for GB.

40. Check It Out! Example 1a
Use the diagram. Find GM.
MZ is a perpendicular bisector of ∆GHJ.
GM = MJ
Circumcenter Thm.
GM = 14.5
Substitute 14.5 for MJ.

41. Check It Out! Example 1b
Use the diagram. Find GK.
KZ is a perpendicular bisector of ∆GHJ.
GK = KH
Circumcenter Thm.
GK = 18.6
Substitute 18.6 for KH.

42. Check It Out! Example 1c
Use the diagram. Find JZ.
Z is the circumcenter of ∆GHJ. By the Circumcenter Theorem, Z is equidistant from the vertices of
∆GHJ.
JZ = GZ
Circumcenter Thm.
JZ = 19.9
Substitute 19.9 for GZ.

43. Example 2: Finding the Circumcenter of a Triangle
Find the circumcenter of ∆HJK with vertices H(0, 0), J(10, 0), and K(0, 6).
Step 1 Graph the triangle.

44. Example 2 Continued
Step 2 Find equations for two perpendicular bisectors.
Since two sides of the triangle lie along the axes, use the graph to find the perpendicular bisectors of these two sides. The perpendicular bisector of HJ is x = 5, and the perpendicular bisector of HK is y = 3.

45. Example 2 Continued
Step 3 Find the intersection of the two equations.
The lines x = 5 and y = 3 intersect at (5, 3), the circumcenter of ∆HJK.

46. Check It Out! Example 2
Find the circumcenter of ∆GOH with vertices G(0, –9), O(0, 0), and H(8, 0) .
Step 1 Graph the triangle.

47. Check It Out! Example 2 Continued
Step 2 Find equations for two perpendicular bisectors.
Since two sides of the triangle lie along the axes, use the graph to find the perpendicular bisectors of these two sides. The perpendicular bisector of GO is y = –4.5, and the perpendicular bisector of OH is
x = 4.

48. Check It Out! Example 2 Continued
Step 3 Find the intersection of the two equations.
The lines x = 4 and y = –4.5 intersect at (4, –4.5), the circumcenter of ∆GOH.

49. A triangle has three angles, so it has three angle bisectors
A triangle has three angles, so it has three angle bisectors. The angle bisectors of a triangle are also concurrent. This point of concurrency is the incenter of the triangle .

50. The distance between a point and a line is the length of the perpendicular segment from the
point to the line.
Remember!

51. Unlike the circumcenter, the incenter is always inside the triangle.

52. The incenter is the center of the triangle’s inscribed circle
The incenter is the center of the triangle’s inscribed circle. A circle inscribed in a polygon intersects each line that contains a side of the polygon at exactly one point.

53. Example 3A: Using Properties of Angle Bisectors
MP and LP are angle bisectors of ∆LMN. Find the distance from P to MN.
P is the incenter of ∆LMN. By the Incenter Theorem, P is equidistant from the sides of ∆LMN.
The distance from P to LM is 5. So the distance from P to MN is also 5.

54. Example 3B: Using Properties of Angle Bisectors
MP and LP are angle bisectors of ∆LMN. Find mPMN.
mMLN = 2mPLN
PL is the bisector of MLN.
mMLN = 2(50°) = 100°
Substitute 50° for mPLN.
mMLN + mLNM + mLMN = 180°
Δ Sum Thm.
mLMN = 180
Substitute the given values.
Subtract 120° from both sides.
mLMN = 60°
PM is the bisector of LMN.
Substitute 60° for mLMN.

55. Check It Out! Example 3a
QX and RX are angle bisectors of ΔPQR. Find the distance from X to PQ.
X is the incenter of ∆PQR. By the Incenter Theorem, X is equidistant from the sides of ∆PQR.
The distance from X to PR is So the distance from X to PQ is also 19.2.

56. QX and RX are angle bisectors of ∆PQR. Find mPQX.
Check It Out! Example 3b
QX and RX are angle bisectors of ∆PQR. Find mPQX.
mQRY= 2mXRY
XR is the bisector of QRY.
mQRY= 2(12°) = 24°
Substitute 12° for mXRY.
mPQR + mQRP + mRPQ = 180°
∆ Sum Thm.
mPQR = 180
Substitute the given values.
Subtract 76° from both sides.
mPQR = 104°
QX is the bisector of PQR.
Substitute 104° for mPQR.

57. Example 4: Community Application
A city planner wants to build a new library between a school, a post office, and a hospital. Draw a sketch to show where the library should be placed so it is the same distance from all three buildings.
Let the three towns be vertices of a triangle. By the Circumcenter Theorem, the circumcenter of the triangle is equidistant from the vertices.
Draw the triangle formed by the three buildings. To find the circumcenter, find the perpendicular bisectors of each side. The position for the library is the circumcenter.

58. Check It Out! Example 4
A city plans to build a firefighters’ monument in the park between three streets. Draw a sketch to show where the city should place the monument so that it is the same distance from all three streets. Justify your sketch.
By the Incenter Thm., the incenter of a ∆ is equidistant from the sides of the ∆. Draw the ∆ formed by the streets and draw the  bisectors to find the incenter, point M. The city should place the monument at point M.

59. Lesson Quiz: Part I
1. ED, FD, and GD are the perpendicular bisectors of ∆ABC. Find BD. 17
2. JP, KP, and HP are angle bisectors of ∆HJK. Find the distance from P to HK. 3

60. Lesson Quiz: Part II
3. Lee’s job requires him to travel to X, Y, and Z. Draw a sketch to show where he should buy a home so it is the same distance from all three
places.