Objectives: Identify properties of perpendicular and angle bisectors
Perpendicular and Angle BisectorsPerpendicular and Angle ......Holt McDougal Geometry 5-1...
Transcript of Perpendicular and Angle BisectorsPerpendicular and Angle ......Holt McDougal Geometry 5-1...
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Holt McDougal Geometry
5-1 Perpendicular and Angle Bisectors 5-1 Perpendicular and Angle Bisectors
Holt Geometry
Warm Up
Lesson Presentation
Lesson Quiz
Holt McDougal Geometry
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Holt McDougal Geometry
5-1 Perpendicular and Angle Bisectors
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).
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Holt McDougal Geometry
5-1 Perpendicular and Angle Bisectors
Prove and apply theorems about
perpendicular bisectors.
Prove and apply theorems about angle
bisectors.
Objectives
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Holt McDougal Geometry
5-1 Perpendicular and Angle Bisectors
equidistant
locus
Vocabulary
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Holt McDougal Geometry
5-1 Perpendicular and Angle Bisectors
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.
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Holt McDougal Geometry
5-1 Perpendicular and Angle Bisectors
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Holt McDougal Geometry
5-1 Perpendicular and Angle Bisectors
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.
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Holt McDougal Geometry
5-1 Perpendicular and Angle Bisectors
Example 1A: Applying the Perpendicular Bisector
Theorem and Its Converse
Find each measure.
MN
MN = LN
MN = 2.6
Bisector Thm.
Substitute 2.6 for LN.
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Holt McDougal Geometry
5-1 Perpendicular and Angle Bisectors
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
BC = 2(12) = 24
Def. of seg. bisector.
Substitute 12 for CD.
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Holt McDougal Geometry
5-1 Perpendicular and Angle Bisectors
Example 1C: Applying the Perpendicular Bisector
Theorem and Its Converse
TU
Find each measure.
So TU = 3(6.5) + 9 = 28.5.
TU = UV Bisector Thm.
3x + 9 = 7x – 17
9 = 4x – 17
26 = 4x
6.5 = x
Subtract 3x from both sides.
Add 17 to both sides.
Divide both sides by 4.
Substitute the given values.
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Holt McDougal Geometry
5-1 Perpendicular and Angle Bisectors
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
DG = 14.6
Bisector Thm.
Substitute 14.6 for EG.
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Holt McDougal Geometry
5-1 Perpendicular and Angle Bisectors
Check It Out! Example 1b
Given that DE = 20.8, DG = 36.4, and EG =36.4, find EF.
Find the measure.
DE = 2EF
20.8 = 2EF
Def. of seg. bisector.
Substitute 20.8 for DE.
Since DG = EG and , is the perpendicular bisector of by the Converse of the Perpendicular Bisector Theorem.
10.4 = EF Divide both sides by 2.
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Holt McDougal Geometry
5-1 Perpendicular and Angle Bisectors
Remember that the distance between a point and a line is the length of the perpendicular segment from the point to the line.
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Holt McDougal Geometry
5-1 Perpendicular and Angle Bisectors
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Holt McDougal Geometry
5-1 Perpendicular and Angle Bisectors
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.
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Holt McDougal Geometry
5-1 Perpendicular and Angle Bisectors
Example 2A: Applying the Angle Bisector Theorem
Find the measure.
BC
BC = DC
BC = 7.2
Bisector Thm.
Substitute 7.2 for DC.
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Holt McDougal Geometry
5-1 Perpendicular and Angle Bisectors
Example 2B: Applying the Angle Bisector Theorem
Find the measure.
mEFH, given that mEFG = 50°.
Since EH = GH,
and , bisects
EFG by the Converse
of the Angle Bisector Theorem.
Def. of bisector
Substitute 50° for mEFG.
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Holt McDougal Geometry
5-1 Perpendicular and Angle Bisectors
Example 2C: Applying the Angle Bisector Theorem
Find mMKL.
, bisects JKL
Since, JM = LM, and
by the Converse of the Angle
Bisector Theorem.
mMKL = mJKM
3a + 20 = 2a + 26
a + 20 = 26
a = 6
Def. of bisector
Substitute the given values.
Subtract 2a from both sides.
Subtract 20 from both sides.
So mMKL = [2(6) + 26]° = 38°
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Holt McDougal Geometry
5-1 Perpendicular and Angle Bisectors
Check It Out! Example 2a
Given that YW bisects XYZ and WZ = 3.05, find WX.
WX = WZ
So WX = 3.05
WX = 3.05
Bisector Thm.
Substitute 3.05 for WZ.
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Holt McDougal Geometry
5-1 Perpendicular and Angle Bisectors
Check It Out! Example 2b
Given that mWYZ = 63°, XW = 5.7, and ZW = 5.7, find mXYZ.
mWYZ = mWYX
mWYZ + mWYX = mXYZ
mWYZ + mWYZ = mXYZ
2(63°) = mXYZ
126° = mXYZ
Bisector Thm.
Substitute m WYZ for
mWYX .
2mWYZ = mXYZ Simplify.
Substitute 63° for mWYZ .
Simplfiy .
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Holt McDougal Geometry
5-1 Perpendicular and Angle Bisectors
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.
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Holt McDougal Geometry
5-1 Perpendicular and Angle Bisectors
Check It Out! Example 3
S is equidistant from each pair of suspension lines. What can you conclude about QS?
QS bisects PQR.
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Holt McDougal Geometry
5-1 Perpendicular and Angle Bisectors
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.
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Holt McDougal Geometry
5-1 Perpendicular and Angle Bisectors
Example 4 Continued
Step 2 Find the midpoint of .
Midpoint formula.
mdpt. of =
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Holt McDougal Geometry
5-1 Perpendicular and Angle Bisectors
Step 3 Find the slope of the perpendicular bisector.
Example 4 Continued
Slope formula.
Since the slopes of perpendicular lines are
opposite reciprocals, the slope of the perpendicular
bisector is
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Holt McDougal Geometry
5-1 Perpendicular and Angle Bisectors
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.
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Holt McDougal Geometry
5-1 Perpendicular and Angle Bisectors
Example 4 Continued
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Holt McDougal Geometry
5-1 Perpendicular and Angle Bisectors
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.
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Holt McDougal Geometry
5-1 Perpendicular and Angle Bisectors
Check It Out! Example 4 Continued
Step 2 Find the midpoint of PQ.
Midpoint formula.
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Holt McDougal Geometry
5-1 Perpendicular and Angle Bisectors
Step 3 Find the slope of the perpendicular bisector.
Check It Out! Example 4 Continued
Slope formula.
Since the slopes of perpendicular lines are
opposite reciprocals, the slope of the
perpendicular bisector is .
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Holt McDougal Geometry
5-1 Perpendicular and Angle Bisectors
Substitute.
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
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Holt McDougal Geometry
5-1 Perpendicular and Angle Bisectors
Lesson Quiz: Part I
Use the diagram for Items 1–2.
1. Given that mABD = 16°, find mABC.
2. Given that mABD = (2x + 12)° and mCBD = (6x – 18)°, find mABC.
32°
54°
65
8.6
Use the diagram for Items 3–4.
3. Given that FH is the perpendicular bisector of EG, EF = 4y – 3, and FG = 6y – 37, find FG.
4. Given that EF = 10.6, EH = 4.3, and FG = 10.6,
find EG.
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Holt McDougal Geometry
5-1 Perpendicular and Angle Bisectors
Lesson Quiz: Part II
5. Write an equation in point-slope form for the perpendicular bisector of the segment with endpoints X(7, 9) and Y(–3, 5) .