1. Midsegment Theorem
The segment connecting the midpoints of two sides of a triangle is parallel to the third side and is half as long as that side. B D E A C

2. In the diagram, UV and VW are midsegments of RST
In the diagram, UV and VW are midsegments of RST. Suppose the distance UW is 80 inches. Find VS.
Answer: 80 inches

3. In the diagram, UV and VW are Midsegments of RST
In the diagram, UV and VW are Midsegments of RST. Suppose the length of RS is 20 inches. Find VW.
Answer: 10 inches

4. In the diagram, UV and VW are midsegments of RST
In the diagram, UV and VW are midsegments of RST. Suppose the distance VU is 82 inches. Find RT.
Answer: 164 inches

5. Applying Variable Coordinates
Graph coordinates P(0, 2), Q(6, 4) and R(4, -2).
How can you find the coordinate of the endpoints of each midsegment of triangle PQR?
Q(6, 4)
Use the Midpoint Formula: • S
P(0, 2) • • • T • V •
R(4, -2)
S = (3, 3)
T = (5, 1)
V = (2, 0)

6. Applying Variable Coordinates
How can you verify that the midsegment theorem is true?
Use the Slope Formula to figure out if the lines are parallel. Q • S • P • • T • V • R

7. Use the Distance Formula of each line segment. .

8. Place a figure in a coordinate plane
Place the figure in a coordinate plane in a way that is convenient for finding side lengths. Assign coordinates to each vertex.
A scalene triangle
It is easy to find lengths of horizontal and vertical segments and distances from (0, 0), so place one vertex at the origin and one or more sides on an axis.

9. Notice that you need to use three different variables.
EXAMPLE 3

10. Apply variable coordinates
Place an isosceles right triangle in a coordinate plane. Then find the length of the hypotenuse and the coordinates of its midpoint M.
Place PQO with the right angle at the origin. Let the length of the legs be k. Then the vertices are located at P(0, k), Q(k, 0), and O(0, 0).

11. Apply variable coordinates
EXAMPLE 4
Apply variable coordinates
Use the Distance Formula to find PQ.
PQ =
(k – 0) + (0 – k) 2 =
k + (– k) 2 =
k + k 2 = 2k 2
= k 2
Use the Midpoint Formula to find the midpoint M of the hypotenuse.
M( )
0 + k , k + 0 2 =
M( , ) k 2

12. EXAMPLE 5
Prove the Midsegment Theorem
Write a coordinate proof of the Midsegment Theorem for one midsegment.
GIVEN :
DE is a midsegment of OBC.
PROVE :
DE OC and DE = OC 1 2 1
Place OBC and assign coordinates. Because you are finding midpoints, use 2p, 2q, and 2r. Then find the coordinates of D and E.
D( )
2q + 0, 2r + 0 2 =
D(q, r)
E( )
2q + 2p, 2r + 0
E(q+p, r)

13. EXAMPLE 5
Prove the Midsegment Theorem 2
Prove DE OC . The y-coordinates of D and E are the same, so DE has a slope of 0. OC is on the x-axis, so its slope is 0.
Because their slopes are the same, DE OC . 3
Prove DE = OC. Use the Ruler Postulate 1 2
to find DE and OC .
DE =
(q + p) – q
= p
OC =
2p – 0
= 2p
So, the length of DE is half the length of OC

14. GUIDED PRACTICE
The example, find the coordinates of F, the midpoint of OC . Then show that EF OB .
(p, 0); slope of EF = = ,
slope of OB = = , the slopes of
EF and OB are both , making EF || OB.
r  0
(q + p)  p q r
2r  0
2q  0
Answer:

15. GUIDED PRACTICE
Graph the points O(0, 0), H(m, n), and J(m, 0). Is OHJ a right triangle? Find the side lengths and the coordinates of the midpoint of each side.
Answer:
yes; OJ = m, JH = n,
HO = m2 + n2,
OJ: ( , 0), JH: (m, ),
HO: ( , ) 2 m n

16. Place the figure in a coordinate plane in a convenient way
Place the figure in a coordinate plane in a convenient way. Assign coordinates to each vertex.
Right triangle: leg lengths are 5 units and 2 units.
(0, 0), (5, 0), (0, 2)

17. Definitions
Perpendicular Bisector: A segment, ray, line, or a plane that is perpendicular to a segment at its midpoint.
Equidistant: a point that is equidistant from two figures if the point is the same distance from each figure.
● A

18. Perpendicular Bisector Theorem:
In a plane, if a point is on the perpendicular bisector of a segment, then it is equidistant from the endpoints.
● C ● A ● B P
If CP is the Perpendicular Bisector of AB, then CA = CB.

19. Converse of the Perpendicular Bisector Theorem
In a plane, if a point is equidistant from the endpoints of a segment, then it is on the perpendicular bisector of the segment.
● D ● A ● B P
● C
If DA = DB, then D lies on the Perpendicular Bisector of AB.

20. Use the Perpendicular Bisector Theorem
BD is the perpendicular bisector of AC Find AD.
AD = CD
Perpendicular Bisector Theorem
3x + 14
5x =
Substitute. 7
x =
Solve for x.
AD =
5x =
5(7) =
35.

21. In the diagram, JK is the perpendicular bisector of NL .1.
What segment lengths are equal?
Find the value of x and NK.
x = 3
NK = 13
In the diagram, JK is the perpendicular bisector of NL , and ML = MN.
8x - 9
1. Find the values of x and y.
4y - 14
x = 4.5
y = 7
2. Find length of NK and MN. 2y 6x
NK = 27
MN = 14

22. Concurrent:
When three or more lines, rays, or segments intersect in the same point.
Point of Concurrency:
Point of intersection of lines, rays or segments.

23. Concurrency of Perpendicular Bisectors of a Triangle
The Perpendicular bisectors of a triangle intersect at a point that is equidistant from the vertices of the triangle. B ●
D ●
● E P ●
A ● ● F
● C
If PD, PE, and PF are perpendicular bisectors, then PA = PB = PC.

24. Circumcenter
The point of concurrency of three perpendicular bisectors of a triangle. P ● P ●

25. Angle Bisector Theorem
If a point is on the bisector of an angle, then it is equidistant from the two sides of the angle. B
26° • D A
26° C
If ray AD bisects angle BAC and BD is perpendicular to AB and DC is perpendicular to AC, then DB = DC.

26. Converse of Angle Bisector Theorem
If a point is in the interior of an angle and is equidistant from the sides of the angle, then it lies on the bisector of the angle. B
26° • D A
26° C
If BD is perpendicular to AB and DC is perpendicular to AC and DB = DC ray AD bisects angle BAC.

27. Concurrency of Angle Bisectors of a Triangle
The angle bisectors of a triangle intersect at a point that is equidistant from the sides of the triangle. B D E ● P A C F
If AP, BP, and CP are angle bisectors of triangle ABC, then PD = PE = PF.

28. Definition
Median of a Triangle – a segment whose endpoints are a vertex of the triangle and the midpoint of the opposite side.

29. Point of Concurrency
The three medians of a triangle are concurrent. The point of concurrency of the medians is called the centroid of the triangle.
The centroid represents the balancing point of the triangle.

30. Theorem 5.8: Concurrency of Medians of a Triangle
The medians of a triangle intersect at a point that is two thirds of the distance from each vertex to the midpoint of the opposite side.
If P is the centroid of  ABC, then

31. Example 1 In RST, Q is the centroid and SQ = 8. Find QW and SW. SQ = 23 SW
Concurrency of Medians of a Triangle Theorem
8 = 2 3 SW
Substitute 8 for SQ.
Multiply each side by the reciprocal, . 3 2
12 = SW
Then QW =
SW – SQ =
12 – 8 = 4.
So, QW = 4 and SW = 12.

32. Example 2
P is the centroid of QRS. PT = 5. Find RT and RP. so

33. Try on your own: C is the centroid of GHJ and CM = 8. Find HM and CH.

34. P is the centroid of triangle ABC, AP = 4, FC = 12, and BF = 9
P is the centroid of triangle ABC, AP = 4, FC = 12, and BF = 9. Find the length of each segment.
AF =
PD =
AD =
AC =
BP =
PF = 12 B 2 6 D E P ● 24 6 A F C 3

35. Definition Acute Triangle
Altitude – the perpendicular segment from a vertex to the opposite side or to the line that contains the opposite side.
Acute Triangle

36. Theorem: Concurrency of Altitudes of a Triangle
The lines containing the altitudes of a triangle are concurrent and intersect at a point called the orthocenter of the triangle.
Altitude
Orthocenter •
Acute Triangle

37. Where is the orthocenter?
Right Triangle A D C B

38. Where is the orthocenter?•
Obtuse Triangle

39. Summary
What is the concurrency point for Perpendicular Bisectors, Angle Bisectors, Medians, and Altitudes of a Triangle?
What is the relationship for Perpendicular Bisectors, Angle Bisectors, and Medians, in a triangle, using the concurrency points?

40. Theorem 5.10
If one side of a triangle is longer than another side, then the angle opposite the longer side is larger than the angle opposite the shorter angle.

41. Theorem 5.11
If one angle of a triangle is larger than another angle, then the side opposite the larger angle is longer than the side opposite the smaller angle.

42. Which side of the Triangle is the longest?
x = 180
120 + x = 180
x = 60
Therefore, since angle A is the largest angle, CB has the longest side.

43. If these inequalities are not true, then you do not have a triangle.
Theorem 5.12
The sum of the lengths of any two sides of a triangle is greater than the length of the third side.
                          
                                         
           
If these inequalities are not true, then you do not have a triangle.

44. Find Possible Side Lengths
5 + x > 9
So, x > 4    
    > x
So, 14 > x     
x + 9 > 5
So, x > -4   
Therefore, 4 < x < 14.

45. Hinge Theorem
If two sides of one triangle are congruent to two sides of another triangle, and the included angle of the first is larger than the included angle of the second then the third side of the first is longer than the third side of the second.

46. Converse of the Hinge Theorem
If two sides of one triangle are congruent to two sides of another triangle, and the third side of the first is longer than the third side of the second then the included angle of the first is larger than the included angle of the second.