Geometry Perpendicular and Angle Bisectors

CONFIDENTIAL 1

Geometry

Perpendicular and Angle Bisectors

CONFIDENTIAL 2

Warm up

Write the equation of each line in slope-intercept from.

1) The line through the points (1,-1) and (2, -9) 2) The line with slope -0.5 through (10, -15) 3) The line with x-intercept -4 and y-intercept 5

CONFIDENTIAL 3

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.


CONFIDENTIAL 4

Distance and Perpendicular BisectorsTheorems

THEOREM HYPOTHESIS CONCLUSION

1.1) Perpendicular Bisector Theorem

If a point is on the perpendicular bisector of a segment, then it is equidistant from the endpoint of the segment.

X

YA B XA = XB

XY AB

YA YB

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CONFIDENTIAL 5


X

YA B

XA = XB

XY AB

YA YB

1.2) Converse of the Perpendicular Bisector TheoremIf a point is equidistant from the endpoints of a segment, then it is on the perpendicular bisector of the segment.

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Perpendicular Bisector Theorem

X

YA B

Given: l is the perpendicular bisector of AB.Prove: XA =XB

Proof: Since l is the perpendicular bisector ofAB, l AB and Y is the midpoint of AB. By thedefinition of perpendicular, AYX and BYXare right angles and AYX BYX. By thedefinition of midpoint, AY BY. By theReflexive Property of Congruence, XY XY.So

AYX BYX by SAS, and XA XB byCPCTC. Therefore XA = XB by the definition ofcongruent segments.

CONFIDENTIAL 7

A locus is a set of point 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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Applying the Perpendicular Bisector Theorem and Its Converse

Find each measure.

W

YX Z

7.3

A) YW

YW = XW YW = 7.3

Bisector Thm.

Substitute 7.3 for XW.

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CONFIDENTIAL 9

D

1636

36

A

B

C

B) BCSince AB = AC and l BC, l is the perpendicularbisector of BC by the Converse of the PerpendicularBisector Theorem.

BC = 2CD

BC = 2(16) = 32

Def. of seg. bisector

Substitute 16 for CD.

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CONFIDENTIAL 10

C) PR

7n - 18

SP Q

R

2n + 9

PR = RQ2n + 9 = 7n – 18 9 = 5n – 18 27 = 5n 5.4 = nSo PR = 2(5.4) + 9 = 19.8

Bisector Thm.

Substitute the given values.

Subtract 2n from both sides.

Add 18 to both sides.

Divide both sides by 5.

CONFIDENTIAL 11

Now you try!

1) Find each measure.

a) Given that line l is the perpendicular bisector of DE and EG = 14.6, find DG.b) Given that DE = 20.8, DG = 36.4, and EG = 36.4, find EF.

G

ED F

CONFIDENTIAL 12

Remember that the distance between a point and a line is the length of the

perpendicular segment from the point to the line.

CONFIDENTIAL 13


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Distance and Angle BisectorsTheorems

A

P

C

B

/APC ≅ /BPC

AC = BC

1.3) Angle Bisector TheoremIf a point is on the bisector of an angle, then it is equidistant from the sides of the angle.

CONFIDENTIAL 14


A

P

C

B

/APC ≅ /BPC

AC = BC

1.4) Converse of the Angle Bisector TheoremIf a point in the interior of an angle is equidistant from the sides of the angle, then it is on the bisector of the angle.

CONFIDENTIAL 15

Based on those 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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Applying the Angle Bisector Theorems

Find each measure.

K

12.8J

M

L

A) LM

LM = JM / Bisector Thm.LM = 12.8 Substitute 12.8 for JM.

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CONFIDENTIAL 17

B) m/ABD, given that m/ABC = 112˚

Since AD = DC, AD BA, andDC BC, BD bisects ABC by the Converse of the Angle BisectorTheorem.mABD =

12 mABC

mABD = 12 (112 ) = 56

Def. of / bisectorSubstitute 112˚ for m/ABC.

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A74

74D

CB

CONFIDENTIAL 18

C) m/TSUU

(6z + 14˚)

(5z + 23˚)S

R

T

Since RU = UT, RU SR, and UT ST, SU bisects RST bythe Converse of the AngleBisector Theorem.

mRSU = mTSU6z + 14 = 5z + 23z + 14 = 23z = 9 so mTUS = [5(9) + 23] = 68

Def. of bisectorSubstitute the given values.Subtract 5z from both sides.Subtract 14 from both sides.

CONFIDENTIAL 19

Now you try!

2) Find each measure.

a) Given that YW bisects XYZ and WZ = 3.05, find WX.

b) Given that m WYZ = 63 , XW = 5.7, and ZW = 5.7, find m XYZ.

W

Y

Z

X

CONFIDENTIAL 20

Parachute Application

PS

R

Q

Each pair of suspension lines on a parachute are the same length and are equally spaced from the center of the chute. How do these lines keep the sky diver centered under the parachute?

It is given that PQ RQ. So Q is on theperpendicular bisector of PR by theConverse of the Perpendicular BisectorTheorem. Since S is the midpoint of PR,QS is the perpendicular bisector of PR.Therefore the sky diver remains centeredunder the chute.

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Now you try!

3)S is equidistant from each pair of suspension lines. What can you conclude about QS?

PS

R

Q

CONFIDENTIAL 22

0 2 4

4

x

y

A

B(1,5)

Writing Equations of Bisectors in Coordinate Plane

Write an equation in point – slope form for the perpendicular bisector of the segment with endpoints A(-1,6) and B(3,4).

Step 1 Graph AB.The Perpendicular bisectorof AB is perpendicular to AB atits midpoint.

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CONFIDENTIAL 23

Step 2 Find the midpoint of AB.x1 + x2

2 , y1 + y2

2 midpoint formula

mdpt. of AB =-1+3

2 ,6+4

2 = (1,5)

0 2 4

4

x

y

A

B(1,5)

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CONFIDENTIAL 24

0 2 4

4

x

y

A

B(1,5)

Step 3 Find the slope of the perpendicular bisector.slope =

y2 - y1

x2 - x1 Slope formula

Slope of AB = 4 - 63-(-1) =

-24 = -

12

Since the slopes of perpendicular lines are oppositereciprocals, the slope of the perpendicular bisectoris 2.

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CONFIDENTIAL 25

0 2 4

4

x

y

A

B(1,5)

Step 4 Use point-slope from to write an eqution. The perpendicular bisector of AB has slope 2 and passes through(1,5). y - y1 = m(x - x1) y - 5 = 2(x - 1)

Point-slope fromSubstitute 5 for y1, 2 for m, and 1 for x1.

CONFIDENTIAL 26

Now you try!

4) Write an equation in point-slope from for the perpendicular bisector of the segment with endpoints

P(5,2) and Q(1,-4).

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Now some problems for you to practice !

CONFIDENTIAL 28

Use the diagram for Exercise.

Assessment

m

S

QP T

a) Given that PS = 53.4, QT = 47.7, and QS = 53.4 find PQ.

b) Given that m is the perpendicular bisector of PQ and SQ = 25.9, find SP.

1)

2)

CONFIDENTIAL 29

Use the diagram for Exercise.

A

D

CB

a) Given that BD bisects ABC and CD = 21.9, find AD.

b) Given that AD = 61, CD = 61, and m ABC = 48 , find mCBD.

3)

4)

CONFIDENTIAL 30

L

M

NP

K

J

5) For a king post truss to be constructed correctly, P must lie on the bisector of /JLK. How can braces PK and PM be used to ensure that P is in the

proper location?

CONFIDENTIAL 31

Write an equation in point-slope from for the perpendicular bisector of the segment with the given

endpoints.

6) M(-5,4), N(1,-2) 7) U(2,-6), V(4,0)

CONFIDENTIAL 32

Let’s review

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.


CONFIDENTIAL 33

Distance and Perpendicular BisectorsTheorems


1.1) Perpendicular Bisector Theorem

If a point is on the perpendicular bisector of a segment, then it is equidistant from the endpoint of the segment.

X

YA B XA = XB

XY AB

YA YB

Next page ->

CONFIDENTIAL 34


X

YA B

XA = XB

XY AB

YA YB

1.2) Converse of the Perpendicular Bisector TheoremIf a point is equidistant from the endpoints of a segment, then it is on the perpendicular bisector of the segment.

CONFIDENTIAL 35

Perpendicular Bisector Theorem

X

YA B

Given: l is the perpendicular bisector of AB.Prove: XA =XB

Proof: Since l is the perpendicular bisector ofAB, l AB and Y is the midpoint of AB. By thedefinition of perpendicular, AYX and BYXare right angles and AYX BYX. By thedefinition of midpoint, AY BY. By theReflexive Property of Congruence, XY XY.So

AYX BYX by SAS, and XA XB byCPCTC. Therefore XA = XB by the definition ofcongruent segments.

CONFIDENTIAL 36

A locus is a set of point 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.

CONFIDENTIAL 37

Applying the Perpendicular Bisector Theorem and Its Converse

Find each measure.

W

YX Z

7.3

A) YW

YW = XW YW = 7.3

Bisector Thm.

Substitute 7.3 for XW.

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CONFIDENTIAL 38

D

1636

36

A

B

C

B) BCSince AB = AC and l BC, l is the perpendicularbisector of BC by the Converse of the PerpendicularBisector Theorem.

BC = 2CD

BC = 2(16) = 32

Def. of seg. bisector

Substitute 16 for CD.

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CONFIDENTIAL 39

C) PR

7n - 18

SP Q

R

2n + 9

PR = RQ2n + 9 = 7n – 18 9 = 5n – 18 27 = 5n 5.4 = nSo PR = 2(5.4) + 9 = 19.8

Bisector Thm.

Substitute the given values.

Subtract 2n from both sides.

Add 18 to both sides.

Divide both sides by 5.

CONFIDENTIAL 40

Remember that the distance between a point and a line is the length of the

perpendicular segment from the point to the line.

CONFIDENTIAL 41


Next page ->

Distance and Angle BisectorsTheorems

A

P

C

B

/APC ≅ /BPC

AC = BC

1.3) Angle Bisector TheoremIf a point is on the bisector of an angle, then it is equidistant from the sides of the angle.

CONFIDENTIAL 42


A

P

C

B

/APC ≅ /BPC

AC = BC

1.4) Converse of the Angle Bisector TheoremIf a point in the interior of an angle is equidistant from the sides of the angle, then it is on the bisector of the angle.

CONFIDENTIAL 43

Based on those 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.

CONFIDENTIAL 44

Applying the Angle Bisector Theorems

Find each measure.

K

12.8J

M

L

A) LM

LM = JM / Bisector Thm.LM = 12.8 Substitute 12.8 for JM.

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CONFIDENTIAL 45

B) m/ABD, given that m/ABC = 112˚

Since AD = DC, AD BA, andDC BC, BD bisects ABC by the Converse of the Angle BisectorTheorem.mABD =

12 mABC

mABD = 12 (112 ) = 56

Def. of / bisectorSubstitute 112˚ for m/ABC.

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A74

74D

CB

CONFIDENTIAL 46

C) m/TSUU

(6z + 14˚)

(5z + 23˚)S

R

T

Since RU = UT, RU SR, and UT ST, SU bisects RST bythe Converse of the AngleBisector Theorem.

mRSU = mTSU6z + 14 = 5z + 23z + 14 = 23z = 9 so mTUS = [5(9) + 23] = 68

Def. of bisectorSubstitute the given values.Subtract 5z from both sides.Subtract 14 from both sides.

CONFIDENTIAL 47

Parachute Application

PS

R

Q

Each pair of suspension lines on a parachute are the same length and are equally spaced from the center of the chute. How do these lines keep the sky diver centered under the parachute?

It is given that PQ RQ. So Q is on theperpendicular bisector of PR by theConverse of the Perpendicular BisectorTheorem. Since S is the midpoint of PR,QS is the perpendicular bisector of PR.Therefore the sky diver remains centeredunder the chute.

CONFIDENTIAL 48

0 2 4

4

x

y

A

B(1,5)

Writing Equations of Bisectors in Coordinate Plane

Write an equation in point – slope form for the perpendicular bisector of the segment with endpoints A(-1,6) and B(3,4).

Step 1 Graph AB.The Perpendicular bisectorof AB is perpendicular to AB atits midpoint.

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CONFIDENTIAL 49

Step 2 Find the midpoint of AB.x1 + x2

2 , y1 + y2

2 midpoint formula

mdpt. of AB =-1+3

2 ,6+4

2 = (1,5)

0 2 4

4

x

y

A

B(1,5)

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CONFIDENTIAL 50

0 2 4

4

x

y

A

B(1,5)

Step 3 Find the slope of the perpendicular bisector.slope =

y2 - y1

x2 - x1 Slope formula

Slope of AB = 4 - 63-(-1) =

-24 = -

12

Since the slopes of perpendicular lines are oppositereciprocals, the slope of the perpendicular bisectoris 2.

Next page ->

CONFIDENTIAL 51

0 2 4

4

x

y

A

B(1,5)

Step 4 Use point-slope from to write an eqution. The perpendicular bisector of AB has slope 2 and passes through(1,5). y - y1 = m(x - x1) y - 5 = 2(x - 1)

Point-slope fromSubstitute 5 for y1, 2 for m, and 1 for x1.

CONFIDENTIAL 52

You did a You did a great great job today!job today!

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Geometry Perpendicular and Angle Bisectors