5.1 Perpendiculars and Bisectors NCSCOS: 2.02&2.03

5.1 Perpendiculars and BisectorsNCSCOS: 2.02&2.03 Geometry Mrs. Spitz Fall 2004

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.

Then and Now • The Alhambra, a place build between 1248 &1254 in Granada, Spain, contains some of the finest examples of Islamic art. The patterns of tiles in the Alhambra fascinated Dutch artist M.C. Escher. Escher built upon the ideas he saw in the tile patterns to create more than 150 designs of repeating patterns that, if continued indefinitely, would fill a plane.

Then and Now continue • In addition to being beautiful pieces of art some of Escher’s patterns have been used to produce unique fabric patterns for clothing and accessories.

Regular division of the plane- Escher Escher’s regular division of plane have become popular themes for scarves, neckties, wrapping paper, and even computer screen background

Escher’s patterns http://www.mcescher.com/Shopmain/ShopEU/facsprints-uk/prints.html

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. Please follow my instructions Use Properties of perpendicular bisectors CP is a  bisector of AB

Perpendicular Bisector Construction – pg. 264 • 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 “class work”

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, line CP  seg. AB and PA = PB, so line CP is the perpendicular bisector of seg. AB.

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.

If a point is on the perpendicular bisector of a segment, then it is equidistant from the endpoints of the segment. If line CP is the perpendicular bisector of segment AB, then CA = CB. Theorem 5.1 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 segment AB. Theorem 5.2: Converse of the Perpendicular Bisector Theorem

Refer to the diagram for Theorem 5.1. 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 Seg. CA ≅ Seg. CB. Exercise 28 asks you to write a two-column proof of Theorem 5.1 using this plan. Plan for Proof of Theorem 5.1

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

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

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 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. Theorem 5.3 Angle Bisector Theorem

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. Ex. 2: Proof of Theorem 4.3

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. Paragraph Proof

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? Ex. 3: Using Angle Bisectors

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. SOLUTION:

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

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 Given: D is in the interior of ABC and is equidistant from BA and BC. Prove: D lies on the angle bisector of ABC. • Reasons: • Given

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 Given: D is in the interior of ABC and is equidistant from BA and BC. Prove: D lies on the angle bisector of ABC. • Reasons: • Given • Given

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 Given: D is in the interior of ABC and is equidistant from BA and BC. Prove: D lies on the angle bisector of ABC. • Reasons: • Given • Given • Def. Equidistant

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 Given: D is in the interior of ABC and is equidistant from BA and BC. Prove: D lies on the angle bisector of ABC. • Reasons: • Given • Given • Def. Equidistant • Def. Distance from point to line.

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 Given: D is in the interior of ABC and is equidistant from BA and BC. Prove: D lies on the angle bisector of ABC. • Reasons: • Given • Given • Def. Equidistant • Def. Distance from point to line. • If 2 lines are , then they form 4 rt. s.

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 Given: D is in the interior of ABC and is equidistant from BA and BC. Prove: D lies on the angle bisector of ABC. • Reasons: • Given • Given • Def. Equidistant • Def. Distance from point to line. • If 2 lines are , then they form 4 rt. s. • Def. of a Right Angle

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 Given: D is in the interior of ABC and is equidistant from BA and BC. Prove: D lies on the angle bisector of ABC. • Reasons: • Given • 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.

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 Given: D is in the interior of ABC and is equidistant from BA and BC. Prove: D lies on the angle bisector of ABC. • Reasons: • Given • 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.

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 Given: D is in the interior of ABC and is equidistant from BA and BC. Prove: D lies on the angle bisector of ABC. • Reasons: • Given • 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

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 Given: D is in the interior of ABC and is equidistant from BA and BC. Prove: D lies on the angle bisector of ABC. • Reasons: • Given • 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.

Assignment: • Check Ms. Vasili’s calendar • Start CHJ#5 • Resource: http://images.search.yahoo.com/search/images?_adv_prop=image&fr=slv8-msgr&va=escher%27s+patterns

5.1 Perpendiculars and Bisectors NCSCOS: 2.02&2.03