Bisectors, Medians, and altitudes in Triangles

1. Bisectors, Medians, and altitudes in Triangles Warm Up Lesson Presentation Lesson Quiz Holt McDougal Geometry Holt Geometry

2. Warm Up 1.Draw a triangle and construct the bisector of one angle. 2.JK is perpendicular to ML at its midpoint K. List the congruent segments.

3. Warm Up 1.What is the name of the point where the angle bisectors of a triangle intersect? Find the midpoint of the segment with the given endpoints. 2. (–1, 6) and (3, 0) 3. (–7, 2) and (–3, –8) 4. Write an equation of the line containing the points (3, 1) and (2, 10) in point-slope form. incenter (1, 3) (–5, –3) y – 1 = –9(x – 3)

4. Objectives Prove and apply properties of perpendicular bisectors of a triangle. Prove and apply properties of angle bisectors of a triangle. Apply properties of medians of a triangle. Apply properties of altitudes of a triangle.

5. Vocabulary concurrent point of concurrency circumcenter of a triangle circumscribed incenter of a triangle inscribed median of a triangle centroid of a triangle altitude of a triangle orthocenter of a triangle

6. Helpful Hint The perpendicular bisector of a side of a triangle does not always pass through the opposite vertex. 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.

7. 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.

8. The circumcenter can be inside the triangle, outside the triangle, or on the triangle.

9. 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.

10. DG, EG, and FG are the perpendicular bisectors of ∆ABC. Find GC. Example 1: Using Properties of Perpendicular Bisectors G is the circumcenter of ∆ABC. By the Circumcenter Theorem, G is equidistant from the vertices of ∆ABC. GC = GB Circumcenter Thm. Substitute 13.4 for GB. GC = 13.4

11. MZ is a perpendicular bisector of ∆GHJ. Check It Out! Example 1a Use the diagram. Find GM. GM = MJ Circumcenter Thm. Substitute 14.5 for MJ. GM = 14.5

12. KZ is a perpendicular bisector of ∆GHJ. Check It Out! Example 1b Use the diagram. Find GK. GK = KH Circumcenter Thm. Substitute 18.6 for KH. GK = 18.6

13. Check It Out! Example 1c 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 Circumcenter Thm. Substitute 19.9 for GZ. JZ = 19.9

14. 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.

15. Unlike the circumcenter, the incenter is always inside the triangle.

16. The incenter is the center of the triangle’s inscribed circle. A circle inscribedin a polygon intersects each line that contains a side of the polygon at exactly one point.

17. MP and LP are angle bisectors of ∆LMN. Find the distance from P to MN. The distance from P to LM is 5. So the distance from P to MN is also 5. Example 3A: Using Properties of Angle Bisectors P is the incenter of ∆LMN. By the Incenter Theorem, P is equidistant from the sides of ∆LMN.

18. PL is the bisector of MLN. PM is the bisector of LMN. Example 3B: Using Properties of Angle Bisectors MP and LP are angle bisectors of ∆LMN. Find mPMN. mMLN = 2mPLN mMLN = 2(50°)= 100° Substitute 50° for mPLN. mMLN + mLNM + mLMN = 180° ΔSum Thm. 100+ 20 + mLMN = 180 Substitute the given values. Subtract 120° from both sides. mLMN = 60° = 30 Substitute 60° for mLMN.

19. QX and RX are angle bisectors of ΔPQR. Find the distance from X to PQ. The distance from X to PR is 19.2. So the distance from X to PQ is also 19.2. Check It Out! Example 3a X is the incenter of ∆PQR. By the Incenter Theorem, X is equidistant from the sides of ∆PQR.

20. XR is the bisector of QRY. QX is the bisector of PQR. Check It Out! Example 3b QX and RX are angle bisectors of ∆PQR. Find mPQX. mQRY= 2mXRY 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. Subtract 76° from both sides. mPQR = 104° Substitute 104° for mPQR.

21. A median of a triangleis a segment whose endpoints are a vertex of the triangle and the midpoint of the opposite side. Every triangle has three medians, and the medians are concurrent.

22. The point of concurrency of the medians of a triangle is the centroid of the triangle. The centroid is always inside the triangle. The centroid is also called the center of gravity because it is the point where a triangular region will balance.

23. Example 1A: Using the Centroid to Find Segment Lengths In ∆LMN, RL = 21 and SQ =4. Find LS. Centroid Thm. Substitute 21 for RL. LS = 14 Simplify.

24. Substitute NQ for NS. Subtract from both sides. Example 1B: Using the Centroid to Find Segment Lengths In ∆LMN, RL = 21 and SQ =4. Find NQ. Centroid Thm. NS + SQ = NQ Seg. Add. Post. Substitute 4 for SQ. 12 = NQ Multiply both sides by 3.

25. Check It Out! Example 1a In ∆JKL, ZW = 7, and LX = 8.1. Find KW. Centroid Thm. Substitute 7 for ZW. KW = 21 Multiply both sides by 3.

26. Check It Out! Example 1b In ∆JKL, ZW = 7, and LX = 8.1. Find LZ. Centroid Thm. Substitute 8.1 for LX. LZ = 5.4 Simplify.

27. An altitude of a triangle is a perpendicular segment from a vertex to the line containing the opposite side. Every triangle has three altitudes. An altitude can be inside, outside, or on the triangle.

28. In ΔQRS, altitude QY is inside the triangle, but RX and SZ are not. Notice that the lines containing the altitudes are concurrent at P. This point of concurrency is the orthocenter of the triangle.