5.1 Midsegments of Triangles Chapter 5 Relationships Within Triangles

5.1 Midsegments of Triangles Theorem 5-1 Triangle Midsegment Theorem If a segment joins the midpoints of two sides of a triangle, then the segment is parallel to the third side, and is half of its length Midsegment x 2x

Midsegment of a Triangle segment whose endpoints are the midpoints of two sides of the triangle Midsegment Base Midsegment = ½ of Base M = ½ b

Find the length of the midsegment. M = ½ b 23 M = ½ • 46 46 M = 23

Find the length of the base. M = ½ b 46 2• 46= ½ b 2• 92 92 = b

Find the length of the midsegment and the base. 5•3-1 = 14 M = ½ b 5x - 1 5x - 1 = ½(6x + 10) 5x - 1 = 3x + 5 6x + 10 -3x -3x 2x - 1 = 5 6•3+10 = 28 +1 +1 2x = 6 2 2 x = 3

5.1 Midsegments of Triangles In ΔEFG, H, J, and K are midpoints. Find HJ, JK, and FG. F HJ: JK: FG: 60 H J 40 G E K 100

5.1 Midsegments of Triangles AB = 10 and CD = 18. Find EB, BC, and AC A B E D C EB: BC: AC:

Example 1: Finding Lengths Theorem 5-1: Triangle Midsegment Theorem If a segment joins the midpoints of two sides of a triangle, then the segment is parallel to the third side, and is half its length Example 1: Finding Lengths In XYZ, M, N and P are the midpoints. The Perimeter of MNP is 60. Find NP and YZ. Because the perimeter is 60, you can find NP. NP + MN + MP = 60 (Definition of Perimeter) NP + + = 60 NP + = 60 NP = 24 22 x P Y M N Z

5.1 Midsegments Find m<VUZ. Justify your answer. 65° X Y V Z U

Example 1 In the diagram, ST and TU are midsegments of triangle PQR. Find PR and TU. 5 ft 16 ft TU = ________ PR = ________

Example 2 In the diagram, XZ and ZY are midsegments of triangle LMN. Find MN and ZY. 14 cm 53 cm ZY = ________ MN = ________

5.1 Midsegments Dean plans to swim the length of the lake, as shown in the photo. He counts the distances shown by counting 3ft strides. How far would Dean swim? 35 strides 118 strides 128 strides x

(5.2) Bisectors in Triangles What will we be learning today? Use properties of perpendicular bisectors and angle bisectors.

If BD = DC, then we say that D is equidistant from B and C. Definition of an Equidistant Point A point D is equidistant from B and C if and only if BD = DC.

T V R M S U Let TU be a perpendicular bisector of RS. RT = TS RV = VS RU = US Then, what can you say about T, V and U? Theorem: If a point lies on the perpendicular bisector of a segment, then the point is equidistant from the endpoints of the segment.

Theorems Theorem 5-2: Perpendicular Bisector Thm. If a point is on the perpendicular bisector of a segment, then it is equidistant form the endpoints of the segment. Theorem 5-3: Converse of the Perpendicular Bisector Thm. If a point is equidistant from the endpoints of a segment, then it is on the perpendicular bisector of the segment.

Conclusion: H lies on the perpendicular bisector of FG. Given: HF = HG Conclusion: H lies on the perpendicular bisector of FG.

the perpendicular bisector of the segment. V R S If T is equidistant from R and S and similarly, V is equidistant from R and S, then what can we say about TV? TV is the perpendicular bisector of RS. Theorem: If two points and a segment lie on the same plane and each of the two points are equidistant from the endpoints of the segment, then the line joining the points is the perpendicular bisector of the segment.

 bisector of a Δ A line, ray, or segment that is  to a side of the Δ at the midpoint of the side. B A M C l

a 90° angles and bisect the side. Perpendicular Bisectors A Perpendicular bisector of a side does not have to start at a vertex. It will form a 90° angles and bisect the side. Circumcenter

Any point on the perpendicular bisector of a segment is equidistance from the endpoints of the segment. AB is the perpendicular bisector of CD A C D B

This makes the Circumcenter an equidistance from the 3 vertices

Theorems Theorem 5-4: Angle Bisector Thm. If a point is on the bisector of an angle, then it is equidistant from the sides of the angle. Theorem 5-5: Converse of the Angle Bisector Thm. If a point in the interior of an angle is equidistant from the sides of the angle, then it is on the angle bisector.

B Let AD be a bisector of BAC, P lie on AD, PM  AB at M, NP  AC at N. M P A D N C Then P is equidistant from AB and AC. Theorem: If a point lies on the bisector of an angle, then the point is equidistant from the sides of the angle.

Key Concepts The distance from a point to a line is the length of the perpendicular segment from the point to the line. Example: D is 3 in. from line AB and line AC C B D A 3

C A B D 5 6

Example Using the Angle Bisector Thm. Find x, FB and FD in the diagram at the right. Show steps to find x, FB and FD: FD = Angle Bisector Thm. 7x – 35 = 2x + 5 7x - 35 A E F C D B 2x + 5

Quick Check a. According to the diagram, how far is K from ray EH? From ray ED? H D C E (X + 20)O 2xO K 10

Quick Check b. What can you conclude about ray EK? 2xO D E C (X + 20)O 10

Quick Check c. Find the value of x. H D C E (X + 20)O 2xO K 10

Quick Check d. Find m<DEH. H D C E (X + 20)O 2xO K 10