1. 5.1 Perpendiculars and Bisectors
Geometry
Ms Reser
Fall

2. Objectives: Use properties of perpendicular bisectors
Use properties of angle bisectors to identify equal distances such as the lengths of beams in a room truss.

3. Assignment:
pp #1-26, 28, all – Quiz after 5.3

4. Use Properties of perpendicular bisectors
In lesson 1.5, you learned that a segment bisector intersects a segment at its midpoint. A segment, ray, line, or plane that is perpendicular to a segment at its midpoint is called a perpendicular bisector. The construction on pg. 264 shows how to draw a line that is perpendicular to a given line or segment at a point P. You can use this method to construct a perpendicular bisector or a segment as described in the activity.
CP is a  bisector of AB

5. Perpendicular Bisector Construction – pg. 264
If you cannot follow these directions, to go page 264 for the instructions with pictures.
Draw a line. Any line about the middle of the page.
Place compass point at P. Draw an arc that intersects line m twice. Label the intersections as A and B.
Use a compass setting greater than AP. Draw an arc from A. With the same setting, draw an arc from B. Label the intersection of the arcs as C.
Use a straightedge to draw CP. This line is perpendicular to line m and passes through P.
Place this in your binder under “computer/lab work”

6. More about perpendicular bisector construction
You can measure CPA on your construction to verify that the constructed line is perpendicular to the given line m. In the construction, CP  AB and PA = PB, so CP is the perpendicular bisector of AB.

7. Equidistant
A point is equidistant from two points if its distance from each point is the same. In the construction above, C is equidistant from A and B because C was drawn so that CA = CB.
Theorem 5.1 states that any point on the perpendicular bisector CP in the construction is equidistant from A and B, the endpoints of the segment. The converse helps you prove that a given point lies on a perpendicular bisector.

8. Theorem 5.1 Perpendicular Bisector Theorem
If a point is on the perpendicular bisector of a segment, then it is equidistant from the endpoints of the segment.
If CP is the perpendicular bisector of AB, then CA = CB.

9. Theorem 5.2: Converse of the Perpendicular Bisector Theorem
If a point is equidistant from the endpoints of a segment, then it is on the perpendicular bisector of the segment.
If DA = DB, then D lies on the perpendicular bisector of AB.

10. Plan for Proof of Theorem 5.1
Refer to the diagram for Theorem Suppose that you are given that CP is the perpendicular bisector of AB. Show that right triangles ∆ABC and ∆BPC are congruent using the SAS Congruence Postulate. Then show that CA ≅ CB.
Exercise 28 asks you to write a two-column proof of Theorem 5.1 using this plan. (This is part of your homework)

11. CP is perpendicular bisector of AB. CP  AB AP ≅ BP CP ≅ CP
Given: CP is perpendicular to AB.
Prove: CA≅CB
Statements:
CP is perpendicular bisector of AB.
CP  AB
AP ≅ BP
CP ≅ CP
CPB ≅ CPA
∆APC ≅ ∆BPC
CA ≅ CB
Reasons:
Given

12. CP is perpendicular bisector of AB. CP  AB AP ≅ BP CP ≅ CP
Given: CP is perpendicular to AB.
Prove: CA≅CB
Statements:
CP is perpendicular bisector of AB.
CP  AB
AP ≅ BP
CP ≅ CP
CPB ≅ CPA
∆APC ≅ ∆BPC
CA ≅ CB
Reasons:
Given
Definition of Perpendicular bisector

13. CP is perpendicular bisector of AB. CP  AB AP ≅ BP CP ≅ CP
Given: CP is perpendicular to AB.
Prove: CA≅CB
Statements:
CP is perpendicular bisector of AB.
CP  AB
AP ≅ BP
CP ≅ CP
CPB ≅ CPA
∆APC ≅ ∆BPC
CA ≅ CB
Reasons:
Given
Definition of Perpendicular bisector

14. CP is perpendicular bisector of AB. CP  AB AP ≅ BP CP ≅ CP
Given: CP is perpendicular to AB.
Prove: CA≅CB
Statements:
CP is perpendicular bisector of AB.
CP  AB
AP ≅ BP
CP ≅ CP
CPB ≅ CPA
∆APC ≅ ∆BPC
CA ≅ CB
Reasons:
Given
Definition of Perpendicular bisector
Reflexive Prop. Congruence.

15. CP is perpendicular bisector of AB. CP  AB AP ≅ BP CP ≅ CP
Given: CP is perpendicular to AB.
Prove: CA≅CB
Statements:
CP is perpendicular bisector of AB.
CP  AB
AP ≅ BP
CP ≅ CP
CPB ≅ CPA
∆APC ≅ ∆BPC
CA ≅ CB
Reasons:
Given
Definition of Perpendicular bisector
Reflexive Prop. Congruence.
Definition right angle

16. CP is perpendicular bisector of AB. CP  AB AP ≅ BP CP ≅ CP
Given: CP is perpendicular to AB.
Prove: CA≅CB
Statements:
CP is perpendicular bisector of AB.
CP  AB
AP ≅ BP
CP ≅ CP
CPB ≅ CPA
∆APC ≅ ∆BPC
CA ≅ CB
Reasons:
Given
Definition of Perpendicular bisector
Reflexive Prop. Congruence.
Definition right angle
SAS Congruence

17. CP is perpendicular bisector of AB. CP  AB AP ≅ BP CP ≅ CP
Given: CP is perpendicular to AB.
Prove: CA≅CB
Statements:
CP is perpendicular bisector of AB.
CP  AB
AP ≅ BP
CP ≅ CP
CPB ≅ CPA
∆APC ≅ ∆BPC
CA ≅ CB
Reasons:
Given
Definition of Perpendicular bisector
Reflexive Prop. Congruence.
Definition right angle
SAS Congruence
CPCTC

18. Ex. 1 Using Perpendicular Bisectors
In the diagram MN is the perpendicular bisector of ST.
What segment lengths in the diagram are equal?
Explain why Q is on MN.

19. Ex. 1 Using Perpendicular Bisectors
What segment lengths in the diagram are equal?
Solution: MN bisects ST, so NS = NT. Because M is on the perpendicular bisector of ST, MS = MT. (By Theorem 5.1). The diagram shows that QS = QT = 12.

20. Ex. 1 Using Perpendicular Bisectors
Explain why Q is on MN.
Solution: QS = QT, so Q is equidistant from S and T. By Theorem 5.2, Q is on the perpendicular bisector of ST, which is MN.

21. Using Properties of Angle Bisectors
The distance from a point to a line is defined as the length of the perpendicular segment from the point to the line. For instance, in the diagram shown, the distance between the point Q and the line m is QP.

22. Using Properties of Angle Bisectors
When a point is the same distance from one line as it is from another line, then the point is equidistant from the two lines (or rays or segments). The theorems in the next few slides show that a point in the interior of an angle is equidistant from the sides of the angle if and only if the point is on the bisector of an angle.

23. Theorem 5.3 Angle Bisector Theorem
If a point is on the bisector of an angle, then it is equidistant from the two sides of the angle.
If mBAD = mCAD, then DB = DC

24. Theorem 5.3 Angle Bisector Theorem
If a point is in the interior of an angle and is equidistant from the sides of the angle, then it lies on the bisector of the angle.
If DB = DC, then mBAD = mCAD.

25. Ex. 2: Proof of Theorem 4.3
Given: D is on the bisector of BAC. DB AB, DC  AC.
Prove: DB = DC
Plan for Proof: Prove that ∆ADB ≅ ∆ADC. Then conclude that DB ≅DC, so DB = DC.

26. Paragraph Proof
By definition of an angle bisector, BAD ≅ CAD. Because ABD and ACD are right angles, ABD ≅ ACD. By the Reflexive Property of Congruence, AD ≅ AD. Then ∆ADB ≅ ∆ADC by the AAS Congruence Theorem. By CPCTC, DB ≅ DC. By the definition of congruent segments DB = DC.

27. Ex. 3: Using Angle Bisectors
Roof Trusses: Some roofs are built with wooden trusses that are assembled in a factory and shipped to the building site. In the diagram of the roof trusses shown, you are given that AB bisects CAD and that ACB and ADB are right angles. What can you say about BC and BD?

28. SOLUTION:
Because BC and BD meet AC and AD at right angles, they are perpendicular segments to the sides of CAD. This implies that their lengths represent distances from the point B to AC and AD. Because point B is on the bisector of CAD, it is equidistant from the sides of the angle.
So, BC = BD, and you can conclude that BC ≅ BD.

29. 32. Developing Proof
Given: D is in the interior of ABC and is equidistant from BA and BC.
Prove: D lies on the angle bisector of ABC.

30. D is in the interior of ABC. D is ___?_ from BA and BC. ____ = ____
Given: D is in the interior of ABC and is equidistant from BA and BC.
Prove: D lies on the angle bisector of ABC.
Statements:
D is in the interior of ABC.
D is ___?_ from BA and BC.
____ = ____
DA  ____, ____  BC
__________
BD ≅ BD
ABD ≅ CBD
BD bisects ABC and point D is on the bisector of ABC
Reasons:
Given

31. Reasons: Given Statements: D is in the interior of ABC.
Given: D is in the interior of ABC and is equidistant from BA and BC.
Prove: D lies on the angle bisector of ABC.
Statements:
D is in the interior of ABC.
D is EQUIDISTANT from BA and BC.
____ = ____
DA  ____, ____  BC
__________
BD ≅ BD
ABD ≅ CBD
BD bisects ABC and point D is on the bisector of ABC
Reasons:
Given

32. Reasons: Statements: D is in the interior of ABC.
Given: D is in the interior of ABC and is equidistant from BA and BC.
Prove: D lies on the angle bisector of ABC.
Statements:
D is in the interior of ABC.
D is EQUIDISTANT from BA and BC.
DA = DC
DA  ____, ____  BC
__________
BD ≅ BD
ABD ≅ CBD
BD bisects ABC and point D is on the bisector of ABC
Reasons:
Given
Def. Equidistant

33. D is in the interior of ABC. D is EQUIDISTANT from BA and BC. DA = DC
Given: D is in the interior of ABC and is equidistant from BA and BC.
Prove: D lies on the angle bisector of ABC.
Statements:
D is in the interior of ABC.
D is EQUIDISTANT from BA and BC.
DA = DC
DA  _BA_, __DC_  BC
__________
BD ≅ BD
ABD ≅ CBD
BD bisects ABC and point D is on the bisector of ABC
Reasons:
Given
Def. Equidistant
Def. Distance from point to line.

34. D is in the interior of ABC. D is EQUIDISTANT from BA and BC. DA = DC
Given: D is in the interior of ABC and is equidistant from BA and BC.
Prove: D lies on the angle bisector of ABC.
Statements:
D is in the interior of ABC.
D is EQUIDISTANT from BA and BC.
DA = DC
DA  _BA_, __DC_  BC
DAB = 90°DCB = 90°
__________
BD ≅ BD
ABD ≅ CBD
BD bisects ABC and point D is on the bisector of ABC
Reasons:
Given
Def. Equidistant
Def. Distance from point to line.
If 2 lines are , then they form 4 rt. s.

35. D is in the interior of ABC. D is EQUIDISTANT from BA and BC. DA = DC
Given: D is in the interior of ABC and is equidistant from BA and BC.
Prove: D lies on the angle bisector of ABC.
Statements:
D is in the interior of ABC.
D is EQUIDISTANT from BA and BC.
DA = DC
DA  _BA_, __DC_  BC
DAB and DCB are rt. s
DAB = 90°DCB = 90°
BD ≅ BD
__________
ABD ≅ CBD
BD bisects ABC and point D is on the bisector of ABC
Reasons:
Given
Def. Equidistant
Def. Distance from point to line.
If 2 lines are , then they form 4 rt. s.
Def. of a Right Angle

36. D is in the interior of ABC. D is EQUIDISTANT from BA and BC. DA = DC
Given: D is in the interior of ABC and is equidistant from BA and BC.
Prove: D lies on the angle bisector of ABC.
Statements:
D is in the interior of ABC.
D is EQUIDISTANT from BA and BC.
DA = DC
DA  _BA_, __DC_  BC
DAB and DCB are rt. s
DAB = 90°DCB = 90°
BD ≅ BD
__________
ABD ≅ CBD
BD bisects ABC and point D is on the bisector of ABC
Reasons:
Given
Def. Equidistant
Def. Distance from point to line.
If 2 lines are , then they form 4 rt. s.
Def. of a Right Angle
Reflexive Property of Cong.

37. D is in the interior of ABC. D is EQUIDISTANT from BA and BC. DA = DC
Given: D is in the interior of ABC and is equidistant from BA and BC.
Prove: D lies on the angle bisector of ABC.
Statements:
D is in the interior of ABC.
D is EQUIDISTANT from BA and BC.
DA = DC
DA  _BA_, __DC_  BC
DAB and DCB are rt. s
DAB = 90°DCB = 90°
BD ≅ BD
∆ABD ≅ ∆CBD
ABD ≅ CBD
BD bisects ABC and point D is on the bisector of ABC
Reasons:
Given
Def. Equidistant
Def. Distance from point to line.
If 2 lines are , then they form 4 rt. s.
Def. of a Right Angle
Reflexive Property of Cong.
HL Congruence Thm.

38. D is in the interior of ABC. D is EQUIDISTANT from BA and BC. DA = DC
Given: D is in the interior of ABC and is equidistant from BA and BC.
Prove: D lies on the angle bisector of ABC.
Statements:
D is in the interior of ABC.
D is EQUIDISTANT from BA and BC.
DA = DC
DA  _BA_, __DC_  BC
DAB and DCB are rt. s
DAB = 90°DCB = 90°
BD ≅ BD
∆ABD ≅ ∆CBD
ABD ≅ CBD
BD bisects ABC and point D is on the bisector of ABC
Reasons:
Given
Def. Equidistant
Def. Distance from point to line.
If 2 lines are , then they form 4 rt. s.
Def. of a Right Angle
Reflexive Property of Cong.
HL Congruence Thm.
CPCTC

39. D is in the interior of ABC. D is EQUIDISTANT from BA and BC. DA = DC
Given: D is in the interior of ABC and is equidistant from BA and BC.
Prove: D lies on the angle bisector of ABC.
Statements:
D is in the interior of ABC.
D is EQUIDISTANT from BA and BC.
DA = DC
DA  _BA_, __DC_  BC
DAB and DCB are rt. s
DAB = 90°DCB = 90°
BD ≅ BD
∆ABD ≅ ∆CBD
ABD ≅ CBD
BD bisects ABC and point D is on the bisector of ABC
Reasons:
Given
Def. Equidistant
Def. Distance from point to line.
If 2 lines are , then they form 4 rt. s.
Def. of a Right Angle
Reflexive Property of Cong.
HL Congruence Thm.
CPCTC
Angle Bisector Thm.