1

Objective – To prove and apply theorems involving angle bisectors and perpendicular bisectors.

Perpendicular Bisector Theorem

Locus - A set of points that satisfies a given condition

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

The locus of points equidistant from the endpoints of a segment.

A B

Perpendicular Bisector -

Given: Line is bisector of ACB is any point on Line Prove: AB BC

nn

Statement1) Draw auxillary lines AB & BC Through 2 pts, is one line2) Line is bisector of ACn

4) AM MC3) M is midpoint of AC

Given

Def. of MidpointDef. of segment bisector

Reasons

A C

B

n M

9) AMB CMB

4) AM MC

7) AMB CMB 6) AMB& CMB are rt. s

Def. of Midpoint

Def. of lines

8) MB MB Reflexive Prop. of

5) Line is to ACn Given

Rt. Thm.

SAS10) AB BC CPCTC11) AB BC Def. of segments

Given: MB is a bisector of ACProve: ABC is isosceles

Statement1) MB is a bisector of AC Given

2) AB BC

A M

B

C

Bisector Thm.

Reasons

)

3) AB BC Def. of segmentsDef. of isosceles 4) ABC is isosceles

M is midpoint of AC

Since AB BC, BM is bisector of AC

Converse of Perpendicular Bisector Theorem

B

Find x.

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

5x 10 x 20 x x

4x 10 20 4x 30x 7.5

M is midpoint of AC

A 5x 10

B

CM

x 20

1010

Angle Bisector Theorem

A

D BD bisects ABC

If a point is in the interior of an angle and it lies on the angle bisector, then it is equidistant from the sides of the angle.

B C

AD DC Since AB BD, CA CD, and AB AC th AD bi t BDC b

Converse of Angle Bisector Theorem

Find m BDA if m CDB =124

B

If a point lies on the interior of an angle and isequidistant from its sides, then it lies on the anglebisector.

AB AC, then AD bisects BDC by Converse of Angle Bisector Theorem.

1m BDA m CDB2

1m BDA (124 )2

A

C

D

m BDA 62

Lesson 5-1

Geometry Slide Show: Teaching Made Easy As Pi, by James Wenk © 2014

2

Write the equation of the perpendicular bisector of thesegment with endpoints C(1,-6) and B(-3,10).

Midpoint

1 2 1 2x x y yM ,

2 2

1 3 6 10M ,2 2

Slope

2 1

2 1

y ym

x x

10 6m3 1

y mx b 1y x b4

12 ( 1) b4

2 4M ,

2 2 1, 2

3 1 16m

4

4 1b 24

1b 24

94

1 9y x4 4

1m4

Write the equation of the perpendicular bisector of thesegment in point-slope form with endpoints A(-4,2) and B(6,-6).

Midpoint

1 2 1 2x x y yM ,

2 2

4 6 2 6M

Slope

2 1

2 1

y ym

x x

6 2

1 1y y m(x x )

1 15y y (x x )4

5m4

1, 2

4 6 2 6M ,2 2

2 4M ,2 2

1, 2

6 2m6 4 8m

10 4

5

1 1y y ( )45y 2 (x 1)4

5m4

Lesson 5-1

Geometry Slide Show: Teaching Made Easy As Pi, by James Wenk © 2014