Pearson Unit 1 Topic 5: Relationships Within Triangles 5-3: Perpendicular and Angle Bisectors Pearson Texas Geometry ©2016 Holt Geometry Texas ©2007

TEKS Focus: (5)(C) Use the constructions of congruent segments, congruent angles, angle bisectors, and perpendicular bisectors to make conjectures about geometric relationships. (1)(E) Create and use representations to organize, record, and communicate mathematical ideas. (1)(F) Analyze mathematical relationships to connect and communicate mathematical ideas. (1)(G) Display, explain, or justify mathematical ideas and arguments using precise mathematical language in written or oral communication. (6)(A) Verify theorems about angles formed by the intersection of lines and line segments, including vertical angles, and angle formed by parallel lines cut by a transversal and prove equidistance between the endpoints of a segment and points on its perpendicular bisector and apply these relationships to solve problems.

Vocabulary: Distance from a point to a line—the length of the perpendicular segment from the point to the line. Equidistant—a point is equidistant from two objects if it is the same distance from the objects. Triangle congruence theorems can be used to prove theorems about equidistant points.

Investigations: INVESTIGATING DISTANCE ON THE PERPENDICULAR BISECTOR OF A SEGMENT INVESTIGATING DISTANCES ON AN ANGLE BISECTOR

Vocabulary: 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.

Vocabulary: The angle bisector of a segment can also be defined as the locus of points in a plane that are equidistant from the sides of the angle.

Example: 1 A park director wants to build a T-shirt stand equidistant from the Rollin’ Coaster and the Spaceship Shoot. What are the possible locations of the stand? Explain. The stand should be built on the perpendicular bisector for the segment between the Rollin’ Coaster and the Spaceship Shoot.

Example: 2 The goalie needs to stand at the point on XY that lies on the angle bisector of angle GPL.

Example: 3 Find MN.  Bisector Thm. MN = LN Substitute 2.6 for LN.

Example: 4 BC = 2(CD) Def. of seg. bisector. BC = 2(12) BC = 24 Find BC. Since AB = AC and , is the perpendicular bisector of by the Converse of the Perpendicular Bisector Theorem. BC = 2(CD) Def. of seg. bisector. BC = 2(12) BC = 24 Substitute 12 for CD.

Example: 5 Find TU. TU = UV  Bisector Thm. 3x + 9 = 7x – 17 Substitute the given values. 9 = 4x – 17 Subtract 3x from both sides. 26 = 4x Add 17 to both sides. 6.5 = x Divide both sides by 4. So TU = 3(6.5) + 9 = 28.5.

Example: 6 Find BC. BC = DC  Bisector Thm. BC = 7.2 Substitute 7.2 for DC.

Example: 7 Find mEFH. Given mEFG = 50°. Def. of  bisector Since EH = GH, and , bisects EFG by the Converse of the Angle Bisector Theorem. Def. of  bisector Substitute 50° for mEFG.

Example: 8 Find mMKL. , bisects JKL Since, JM = LM, and by the Converse of the Angle Bisector Theorem. mMKL = mJKM Def. of  bisector 3a + 20 = 2a + 26 Substitute the given values. a + 20 = 26 Subtract 2a from both sides. a = 6 Subtract 20 from both sides. So mMKL = [2(6) + 26]° = 38°

Example: 9 Given that YW bisects XYZ and WZ = 3.05, find WX. WX = WZ  Bisector Thm. WX = 3.05 Substitute 3.05 for WZ. So WX = 3.05

Example: 10 Given that mWYZ = 63°, XW = 5.7, and ZW = 5.7, find mXYZ.  Addition Postulate mWYZ + mWYX = mXYZ  Bisector Thm. mWYZ = mWYX Substitute m WYZ for mWYX . mWYZ + mWYZ = mXYZ 2mWYZ = mXYZ Simplify. 2(63°) = mXYZ Substitute 63° for mWYZ . 126° = mXYZ Simplfiy .

Example: 11 John wants to hang a spotlight along the back of a display case. Wires AD and CD are the same length, and points A and C are equidistant from B. How do the wires keep the spotlight centered? It is given that . So B is on the angle bisector of ADC 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.

Example: 12 S is equidistant from each pair of suspension lines. What can you conclude about QS? QS bisects  PQR.

Example: 13 Write an equation in point-slope form for the perpendicular bisector of the segment with endpoints M(-3, -1) and N(7, –5). Step 1 Graph PQ. The perpendicular bisector of is perpendicular to at its midpoint.

Example: 13 continued −3+7 2 , −1+(−5) 2 = 4 2 , −6 2 = 2, −3 Step 2 Find the midpoint of PQ. Midpoint formula. −3+7 2 , −1+(−5) 2 = 4 2 , −6 2 = 2, −3 Step 3 Find the slope of the perpendicular bisector. Slope formula. −5 −(−1) 7 −(−3) = −5+1 7+3 = −4 10 = −2 5 Since the slopes of perpendicular lines are opposite reciprocals, the slope of the perpendicular bisector is 5/2 .

Example: 13 continued Step 4 Use point-slope form to write an equation. The perpendicular bisector of PQ has slope 5/2 and passes through (2, -3). y – y1 = m(x – x1) Point-slope form y – (-3) = 5/2(x – 2) Substitute y + 3 = 5/2(x – 2) Simplify Now change to slope-intercept form y + 3 = (5/2)x – 5 Distribute y = (5/2)x – 8 Subtract 3