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Geometry 201 unit 5.3
This document discusses bisectors in triangles. It defines perpendicular bisectors as lines that bisect and are perpendicular to a side of a triangle. Angle bisectors bisect an angle of a triangle. The three perpendicular bisectors and angle bisectors of a triangle are both concurrent, meaning they intersect at a single point. The point of concurrency of the perpendicular bisectors is called the circumcenter and is equidistant from the triangle's vertices. The point of concurrency of the angle bisectors is called the incenter and is always inside the triangle, equidistant from its sides. Examples show using bisectors to find distances and angles in triangles, as well as applications like placing a building or monument equidistant from three
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Geometry 201 unit 5.3
- 1. UNIT 5.3 BISECTORSINUNIT 5.3 BISECTORS IN A TRIANGLEA TRIANGLE
- 2. Warm Up 1. Drawa triangle and construct the bisector of one angle. 2. JK is perpendicular to ML at its midpoint K. List the congruent segments.
- 3. Prove and applyproperties of perpendicular bisectors of a triangle. Prove and apply properties of angle bisectors of a triangle. Objectives
- 4. concurrent point of concurrency circumcenterof a triangle circumscribed incenter of a triangle inscribed Vocabulary
- 5. Since a trianglehas three sides, it has three perpendicular bisectors. When you construct the perpendicular bisectors, you find that they have an interesting property.
- 6. The perpendicular bisectorof a side of a triangle does not always pass through the opposite vertex. Helpful Hint
- 7. When three ormore 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.
- 8. The circumcenter canbe inside the triangle, outside the triangle, or on the triangle.
- 9. The circumcenter ofΔABC is the center of its circumscribed circle. A circle that contains all the vertices of a polygon is circumscribed about the polygon.
- 10. Example 1: UsingProperties of Perpendicular Bisectors G is the circumcenter of ∆ABC. By the Circumcenter Theorem, G is equidistant from the vertices of ∆ABC. DG, EG, and FG are the perpendicular bisectors of ∆ABC. Find GC. GC = CB GC = 13.4 Circumcenter Thm. Substitute 13.4 for GB.
- 11. Check It Out!Example 1a Use the diagram. Find GM. GM = MJ GM = 14.5 Circumcenter Thm. Substitute 14.5 for MJ. MZ is a perpendicular bisector of ∆GHJ.
- 12. Check It Out!Example 1b Use the diagram. Find GK. GK = KH GK = 18.6 Circumcenter Thm. Substitute 18.6 for KH. KZ is a perpendicular bisector of ∆GHJ.
- 13. Check It Out!Example 1c Use the diagram. Find JZ. JZ = GZ JZ = 19.9 Circumcenter Thm. Substitute 19.9 for GZ. Z is the circumcenter of ∆GHJ. By the Circumcenter Theorem, Z is equidistant from the vertices of ∆GHJ.
- 14. Example 2: Findingthe 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.
- 15. Example 2 Continued Step2 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.
- 16. Example 2 Continued Step3 Find the intersection of the two equations. The lines x = 5 and y = 3 intersect at (5, 3), the circumcenter of ∆HJK.
- 17. 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.
- 18. 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.
- 19. 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.
- 20. A triangle hasthree 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 .
- 21. The distance betweena point and a line is the length of the perpendicular segment from the point to the line. Remember!
- 22. Unlike the circumcenter,the incenter is always inside the triangle.
- 23. The incenter isthe 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.
- 24. Example 3A: UsingProperties 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.
- 25. Example 3B: UsingProperties of Angle Bisectors MP and LP are angle bisectors of ∆LMN. Find m∠PMN. m∠MLN = 2m∠PLN m∠MLN = 2(50°) = 100° m∠MLN + m∠LNM + m∠LMN = 180° 100 + 20 + m∠LMN = 180 m∠LMN = 60° Substitute 50° for m∠PLN. Δ Sum Thm. Substitute the given values. Subtract 120° from both sides. Substitute 60° for m∠LMN. PL is the bisector of ∠MLN. PM is the bisector of ∠LMN.
- 26. Check It Out!Example 3a 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.
- 27. QX and RXare angle bisectors of ∆PQR. Find m∠PQX. m∠QRY= 2m∠XRY m∠QRY= 2(12°) = 24° m∠PQR + m∠QRP + m∠RPQ = 180° m∠PQR + 24 + 52 = 180 m∠PQR = 104° Substitute 12° for m∠XRY. ∆ Sum Thm. Substitute the given values. Subtract 76° from both sides. Substitute 104° for m∠PQR. XR is the bisector of ∠QRY. QX is the bisector of ∠PQR. Check It Out! Example 3b
- 28. Example 4: CommunityApplication 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.
- 29. 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.
- 30. Lesson Quiz: PartI 1. ED, FD, and GD are the perpendicular bisectors of ∆ABC. Find BD. 17 3 2. JP, KP, and HP are angle bisectors of ∆HJK. Find the distance from P to HK.
- 31. Lesson Quiz: PartII 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.
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