and Angle Bisectors 5 1 Perpendicular and Angle
and Angle Bisectors 5 -1 Perpendicular and Angle Bisectors Warm Up Lesson Presentation Lesson Quiz Holt Geometry
5 -1 Perpendicular and Angle Bisectors Warm Up Construct each of the following. 1. A perpendicular bisector. 2. An angle bisector. 3. Find the midpoint and slope of the segment (2, 8) and (– 4, 6). Holt Geometry
5 -1 Perpendicular and Angle Bisectors Objectives Prove and apply theorems about perpendicular bisectors. Prove and apply theorems about angle bisectors. Holt Geometry
5 -1 Perpendicular and Angle Bisectors Vocabulary equidistant locus Holt Geometry
5 -1 Perpendicular and Angle Bisectors 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. Holt Geometry
5 -1 Perpendicular and Angle Bisectors Holt Geometry
5 -1 Perpendicular and Angle Bisectors 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. Holt Geometry
5 -1 Perpendicular and Angle Bisectors Example 1 A: Applying the Perpendicular Bisector Theorem and Its Converse Find each measure. MN MN = LN Bisector Thm. MN = 2. 6 Substitute 2. 6 for LN. Holt Geometry
5 -1 Perpendicular and Angle Bisectors Example 1 B: Applying the Perpendicular Bisector Theorem and Its Converse Find each measure. 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) = 24 Substitute 12 for CD. Holt Geometry
5 -1 Perpendicular and Angle Bisectors Example 1 C: Applying the Perpendicular Bisector Theorem and Its Converse Find each measure. TU TU = UV Bisector Thm. 3 x + 9 = 7 x – 17 Substitute the given values. 9 = 4 x – 17 Subtract 3 x from both sides. 26 = 4 x 6. 5 = x Add 17 to both sides. Divide both sides by 4. So TU = 3(6. 5) + 9 = 28. 5. Holt Geometry
5 -1 Perpendicular and Angle Bisectors Check It Out! Example 1 a Find the measure. Given that line ℓ is the perpendicular bisector of DE and EG = 14. 6, find DG. DG = EG Bisector Thm. DG = 14. 6 Substitute 14. 6 for EG. Holt Geometry
5 -1 Perpendicular and Angle Bisectors Check It Out! Example 1 b Find the measure. Given that DE = 20. 8, DG = 36. 4, and EG =36. 4, find EF. Since DG = EG and , is the perpendicular bisector of by the Converse of the Perpendicular Bisector Theorem. DE = 2 EF Def. of seg. bisector. 20. 8 = 2 EF Substitute 20. 8 for DE. 10. 4 = EF Divide both sides by 2. Holt Geometry
5 -1 Perpendicular and Angle Bisectors Remember that the distance between a point and a line is the length of the perpendicular segment from the point to the line. Holt Geometry
5 -1 Perpendicular and Angle Bisectors Holt Geometry
5 -1 Perpendicular and Angle Bisectors Based on these 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. Holt Geometry
5 -1 Perpendicular and Angle Bisectors Example 2 A: Applying the Angle Bisector Theorem Find the measure. BC BC = DC Bisector Thm. BC = 7. 2 Substitute 7. 2 for DC. Holt Geometry
5 -1 Perpendicular and Angle Bisectors Example 2 B: Applying the Angle Bisector Theorem Find the measure. m EFH, given that m EFG = 50°. Since EH = GH, and , bisects EFG by the Converse of the Angle Bisector Theorem. Def. of bisector Substitute 50° for m EFG. Holt Geometry
5 -1 Perpendicular and Angle Bisectors Example 2 C: Applying the Angle Bisector Theorem Find m MKL. Since, JM = LM, and , bisects JKL by the Converse of the Angle Bisector Theorem. m MKL = m JKM Def. of bisector 3 a + 20 = 2 a + 26 a + 20 = 26 a=6 Substitute the given values. Subtract 2 a from both sides. Subtract 20 from both sides. So m MKL = [2(6) + 26]° = 38° Holt Geometry
5 -1 Perpendicular and Angle Bisectors Check It Out! Example 2 a 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 Holt Geometry
5 -1 Perpendicular and Angle Bisectors Check It Out! Example 2 b Given that m WYZ = 63°, XW = 5. 7, and ZW = 5. 7, find m XYZ. m WYZ + m WYX = m XYZ m WYZ = m WYX m WYZ + m WYZ = m XYZ 2(63°) = m XYZ 126° = m XYZ Holt Geometry Bisector Thm. Substitute m WYZ for m WYX. Simplify. Substitute 63° for m WYZ. Simplfiy.
5 -1 Perpendicular and Angle Bisectors Example 3: Application John wants to hang a spotlight along the back of a display case. Wires AD and CD are the same length, and A and C are equidistant from B. How do the wires keep the spotlight centered? It is given that. So D is on the perpendicular bisector of 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. Holt Geometry
5 -1 Perpendicular and Angle Bisectors Check It Out! Example 3 S is equidistant from each pair of suspension lines. What can you conclude about QS? QS bisects PQR. Holt Geometry
5 -1 Perpendicular and Angle Bisectors Example 4: Writing Equations of Bisectors in the Coordinate Plane Write an equation in point-slope form for the perpendicular bisector of the segment with endpoints C(6, – 5) and D(10, 1). Step 1 Graph . The perpendicular bisector of is perpendicular to at its midpoint. Holt Geometry
5 -1 Perpendicular and Angle Bisectors Example 4 Continued Step 2 Find the midpoint of . Midpoint formula. mdpt. of Holt Geometry =
5 -1 Perpendicular and Angle Bisectors Example 4 Continued Step 3 Find the slope of the perpendicular bisector. Slope formula. Since the slopes of perpendicular lines are opposite reciprocals, the slope of the perpendicular bisector is Holt Geometry
5 -1 Perpendicular and Angle Bisectors Example 4 Continued Step 4 Use point-slope form to write an equation. The perpendicular bisector of has slope and passes through (8, – 2). y – y 1 = m(x – x 1) Point-slope form Substitute – 2 for y 1, for x 1. Holt Geometry for m, and 8
5 -1 Perpendicular and Angle Bisectors Example 4 Continued Holt Geometry
5 -1 Perpendicular and Angle Bisectors Check It Out! Example 4 Write an equation in point-slope form for the perpendicular bisector of the segment with endpoints P(5, 2) and Q(1, – 4). Step 1 Graph PQ. The perpendicular bisector of is perpendicular to at its midpoint. Holt Geometry
5 -1 Perpendicular and Angle Bisectors Check It Out! Example 4 Continued Step 2 Find the midpoint of PQ. Midpoint formula. Holt Geometry
5 -1 Perpendicular and Angle Bisectors Check It Out! Example 4 Continued Step 3 Find the slope of the perpendicular bisector. Slope formula. Since the slopes of perpendicular lines are opposite reciprocals, the slope of the perpendicular bisector is Holt Geometry .
5 -1 Perpendicular and Angle Bisectors Check It Out! Example 4 Continued Step 4 Use point-slope form to write an equation. The perpendicular bisector of PQ has slope passes through (3, – 1). y – y 1 = m(x – x 1) Point-slope form Substitute. Holt Geometry and
5 -2 Bisectors of Triangles Objectives Prove and apply properties of perpendicular bisectors of a triangle. Prove and apply properties of angle bisectors of a triangle. Holt Geometry
5 -2 Bisectors of Triangles Vocabulary concurrent point of concurrency circumcenter of a triangle circumscribed incenter of a triangle inscribed Holt Geometry
5 -2 Bisectors of Triangles Since a triangle has three sides, it has three perpendicular bisectors. When you construct the perpendicular bisectors, you find that they have an interesting property. Holt Geometry
5 -2 Bisectors of Triangles Helpful Hint The perpendicular bisector of a side of a triangle does not always pass through the opposite vertex. Holt Geometry
5 -2 Bisectors of Triangles When three or more lines intersect at one point, the lines are said to be concurrent. The point of concurrency is the point where they intersect. In the construction, you saw that the three perpendicular bisectors of a triangle are concurrent. This point of concurrency is the circumcenter of the triangle. Holt Geometry
5 -2 Bisectors of Triangles The circumcenter can be inside the triangle, outside the triangle, or on the triangle. Holt Geometry
5 -2 Bisectors of Triangles The circumcenter of ΔABC is the center of its circumscribed circle. A circle that contains all the vertices of a polygon is circumscribed about the polygon. Holt Geometry
5 -2 Bisectors of Triangles Example 1: Using Properties of Perpendicular Bisectors DG, EG, and FG are the perpendicular bisectors of ∆ABC. Find GC. G is the circumcenter of ∆ABC. By the Circumcenter Theorem, G is equidistant from the vertices of ∆ABC. GC = CB GC = 13. 4 Holt Geometry Circumcenter Thm. Substitute 13. 4 for GB.
5 -2 Bisectors of Triangles Check It Out! Example 1 a Use the diagram. Find GM. MZ is a perpendicular bisector of ∆GHJ. GM = MJ GM = 14. 5 Holt Geometry Circumcenter Thm. Substitute 14. 5 for MJ.
5 -2 Bisectors of Triangles Check It Out! Example 1 b Use the diagram. Find GK. KZ is a perpendicular bisector of ∆GHJ. GK = KH GK = 18. 6 Holt Geometry Circumcenter Thm. Substitute 18. 6 for KH.
5 -2 Bisectors of Triangles Check It Out! Example 1 c Use the diagram. Find JZ. Z is the circumcenter of ∆GHJ. By the Circumcenter Theorem, Z is equidistant from the vertices of ∆GHJ. JZ = GZ JZ = 19. 9 Holt Geometry Circumcenter Thm. Substitute 19. 9 for GZ.
5 -2 Bisectors of Triangles Example 2: Finding the Circumcenter of a Triangle Find the circumcenter of ∆HJK with vertices H(0, 0), J(10, 0), and K(0, 6). Step 1 Graph the triangle. Holt Geometry
5 -2 Bisectors of Triangles Example 2 Continued Step 2 Find equations for two perpendicular bisectors. Since two sides of the triangle lie along the axes, use the graph to find the perpendicular bisectors of these two sides. The perpendicular bisector of HJ is x = 5, and the perpendicular bisector of HK is y = 3. Holt Geometry
5 -2 Bisectors of Triangles Example 2 Continued Step 3 Find the intersection of the two equations. The lines x = 5 and y = 3 intersect at (5, 3), the circumcenter of ∆HJK. Holt Geometry
5 -2 Bisectors of Triangles Check It Out! Example 2 Find the circumcenter of ∆GOH with vertices G(0, – 9), O(0, 0), and H(8, 0). Step 1 Graph the triangle. Holt Geometry
5 -2 Bisectors of Triangles Check It Out! Example 2 Continued Step 2 Find equations for two perpendicular bisectors. Since two sides of the triangle lie along the axes, use the graph to find the perpendicular bisectors of these two sides. The perpendicular bisector of GO is y = – 4. 5, and the perpendicular bisector of OH is x = 4. Holt Geometry
5 -2 Bisectors of Triangles Check It Out! Example 2 Continued Step 3 Find the intersection of the two equations. The lines x = 4 and y = – 4. 5 intersect at (4, – 4. 5), the circumcenter of ∆GOH. Holt Geometry
5 -2 Bisectors of Triangles A triangle has three angles, so it has three angle bisectors. The angle bisectors of a triangle are also concurrent. This point of concurrency is the incenter of the triangle. Holt Geometry
5 -2 Bisectors of Triangles Remember! The distance between a point and a line is the length of the perpendicular segment from the point to the line. Holt Geometry
5 -2 Bisectors of Triangles Unlike the circumcenter, the incenter is always inside the triangle. Holt Geometry
5 -2 Bisectors of Triangles The incenter is the center of the triangle’s inscribed circle. A circle inscribed in a polygon intersects each line that contains a side of the polygon at exactly one point. Holt Geometry
5 -2 Bisectors of Triangles Example 3 A: Using Properties of Angle Bisectors MP and LP are angle bisectors of ∆LMN. Find the distance from P to MN. P is the incenter of ∆LMN. By the Incenter Theorem, P is equidistant from the sides of ∆LMN. The distance from P to LM is 5. So the distance from P to MN is also 5. Holt Geometry
5 -2 Bisectors of Triangles Example 3 B: Using Properties of Angle Bisectors MP and LP are angle bisectors of ∆LMN. Find m PMN. m MLN = 2 m PLN PL is the bisector of MLN. m MLN = 2(50°) = 100° Substitute 50° for m PLN. m MLN + m LNM + m LMN = 180° Δ Sum Thm. 100 + 20 + m LMN = 180 Substitute the given values. m LMN = 60° Subtract 120° from both sides. PM is the bisector of LMN. Substitute 60° for m LMN. Holt Geometry
5 -2 Bisectors of Triangles Check It Out! Example 3 a QX and RX are angle bisectors of ΔPQR. Find the distance from X to PQ. X is the incenter of ∆PQR. By the Incenter Theorem, X is equidistant from the sides of ∆PQR. The distance from X to PR is 19. 2. So the distance from X to PQ is also 19. 2. Holt Geometry
5 -2 Bisectors of Triangles Check It Out! Example 3 b QX and RX are angle bisectors of ∆PQR. Find m PQX. m QRY= 2 m XRY XR is the bisector of QRY. m QRY= 2(12°) = 24° Substitute 12° for m XRY. m PQR + m QRP + m RPQ = 180° ∆ Sum Thm. m PQR + 24 + 52 = 180 Substitute the given values. m PQR = Subtract 76° from both 104° sides. QX is the bisector of PQR. Substitute 104° for m PQR. Holt Geometry
5 -2 Bisectors of Triangles Example 4: Community Application A city planner wants to build a new library between a school, a post office, and a hospital. Draw a sketch to show where the library should be placed so it is the same distance from all three buildings. Let the three towns be vertices of a triangle. By the Circumcenter Theorem, the circumcenter of the triangle is equidistant from the vertices. Draw the triangle formed by the three buildings. To find the circumcenter, find the perpendicular bisectors of each side. The position for the library is the circumcenter. Holt Geometry
5 -2 Bisectors of Triangles Check It Out! Example 4 A city plans to build a firefighters’ monument in the park between three streets. Draw a sketch to show where the city should place the monument so that it is the same distance from all three streets. Justify your sketch. By the Incenter Thm. , the incenter of a ∆ is equidistant from the sides of the ∆. Draw the ∆ formed by the streets and draw the bisectors to find the incenter, point M. The city should place the monument at point M. Holt Geometry
5 -2 Bisectors of Triangles Lesson Quiz: Part I 1. ED, FD, and GD are the perpendicular bisectors of ∆ABC. Find BD. 17 2. JP, KP, and HP are angle bisectors of ∆HJK. Find the distance from P to HK. 3 Holt Geometry
5 -2 Bisectors of Triangles Lesson Quiz: Part II 3. Lee’s job requires him to travel to X, Y, and Z. Draw a sketch to show where he should buy a home so it is the same distance from all three places. Holt Geometry
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