1. Special Segments in Triangles
Triangle Properties
Special Segments in Triangles
A line segment consists of two points called the endpoints of the segment and all the points between them that are collinear with the two points.
Two segments are congruent if and only if they have equal measures, or lengths.
The midpoint of a segment is the point on the segment that is the same distance from both endpoints. The midpoint bisects the segment, or divides the segment into two congruent segments. Each segment has exactly one midpoint.
A segment bisector is a line, ray or segment that passes through the midpoint of a segment.

2. Use Perpendicular Bisectors
Perpendicular bisector – a segment, ray, line, or plane that is perpendicular to a segment at its midpoint
Equidistant - same distance
Circumcenter – the point of concurrency of the 3 perpendicular bisectors of a triangle

3. Perpendicular Bisector Theorem
If a point is on the perpendicular bisector of a segment, then it is equidistant from the endpoints of the segment.
If CP is the perp bisector of AB, then CA = CB
Converse of the Perpendicular Bisector Theorem
If a point is equidistant from the endpoints of the segment, then it is on the perpendicular bisector of the segment.
If DA = DB, then D lies on the CP.
perp bisector

4. Perpendicular Bisector Theorem
If AC = BC, then C A B
C is on the perpendicular bisector of AB.

5. Perpendicular Bisector Theorem
If l is the perpendicular bisector of AB,
then AC = BC and AD = BD.

6. Perpendicular Bisector – Special Segment of a triangle
A line (or ray or segment) that is perpendicular to a segment at its midpoint.
Definition:
The perpendicular bisector does not have to start from a vertex! R O Q P
Example: M L N C D A E A B B
In the isosceles ∆POQ, is the perpendicular bisector.
In the scalene ∆CDE, is the perpendicular bisector.
In the right ∆MLN, is the perpendicular bisector.

7. JM = ML
 Bisector Thm
7x = 3x + 16
Substitute
– 3x = – 3x
Solve for x
4x = 16
4 = 4
x = 4
ML = 3x + 16
= 3(4) + 16 =
= 28

8. AC = CB
DC bisects AB
AE = BE
Definition of Congruence
AD = BD
Bisector Thm.
Yes, because AE = BE, E is equidistant from B and A. According to Converse of ⊥ Bisector, E is on the ⊥ Bisector DC.

9. KG = KH
, JG = JH
, FG = FH
KG = KH
GH = KG + KH
2x = x + 1
GH = 2x + (x + 1)
-x -x
GH = 2(1)+ (1 + 1)
GH = 2+ (2)
x = 1
GH = 4

10. Perpendicular Bisectors of a Triangle Every triangle has 3 perpendicular bisectors.

11. Perpendicular Bisectors of a Triangle

12. Perpendicular Bisectors of a Triangle

13. The 3 perpendicular bisectors of any triangle will intersect at a point that is equidistant from the vertices of the triangle. This point is called the circumcenter and is the center of a circle that contains all 3 vertices of the triangle.

14. Median of a Triangle
Median of a Triangle: A segment is a median of a triangle if one endpoint is a vertex of the triangle and the other endpoint is the midpoint of the side opposite that vertex.
Midsegment: The segment that connects the midpoints of two sides of a triangle. A triangle has three sides, each with its own midpoint, so there are three midsegments in every triangle.

16. Median - Special Segment of Triangle
Definition:
A segment from the vertex of the triangle to the midpoint of the opposite side. B A D E C F
Since there are three vertices, there are three medians.
In the figure C, E and F are the midpoints of the sides of the triangle.

17. Medians of a Triangle
Every triangle has 3 medians.

18. Medians of a Triangle
Every triangle has 3 medians.

19. Medians of a Triangle
Every triangle has 3 medians.

20. Centroids
The medians of a triangle will always intersect at the same point - the centroid. The centroid of a triangle is located 2/3 of the distance from the vertex to the midpoint of the opposite side.
The centroid is always inside the triangle
Each median divides the triangle into two smaller triangles of equal area.
Put another way, the centroid divides each median into two segments whose lengths are in the ratio 2:1, with the longest one nearest the vertex.

21. Centroid
centroid

22. Centroid

23. Concurrency of Medians of a Triangle (Centroid)
The medians of a intersect at a point that is two thirds of the distance from each vertex to the midpoint of the opposite side.
In other words, the distance from the vertex to the centroid is twice the distance from the centroid to the midpoint.
AB = AC + CB
If AC is 2/3 of AB, what is CB?
CB = 1/3
If AB = 9, what is AC and CB?
AC = 6
What do you notice about AC and CB?
AC is twice CB
Why?
Now assume, A is vertex, C is centroid, and B is midpoint of opposite side.
Vertex to centroid = 2/3 median
CB = 3
Centroid to midpoint = 1/3 median

24. P is the centroid of ABC. What relationships exist? AP = 2  PE
The dist. from the vertex to the centroid is twice the dist. from centroid to midpoint.
AP = 2  PE
BP = 2  PF
CP = 2  PD
The dist. from the centroid to midpoint is half the dist. from the vertex to the centroid.
PE = ½  AP
PF = ½  BP
PD = ½  CP
The dist. from the vertex to the centroid is 2/3 the distance of the median.
AP = 2/3  AE
BP = 2/3  BF
CP = 2/3  CD
The dist. from the centroid to midpoint is 1/3 the distance of the median.
PE = 1/3  AE
PF = 1/3  BF
PD = 1/3  CD

25. BG = 12 + 6 BG = 18 What do I know about DG? What do I know about BG?
BG = BD + DG
BG =
DG = 6
BG = 18

26. Your Turn
In PQR, S is the centroid, UQ = 5, TR = 3, RV = 5, and SU = 2.
1. Find RU and RS.
RU is a median and RU = RS + SU.
RS = 2  SU
RU = RS + SU
RU = 4 + 2
RS = 2  2
RU = 6
RS = 4
2. Find the perimeter of PQR.
Perimeter means add up the sides of the triangle.
U is midpoint of PQ so PU = UQ , PU = 5
QT, RU, and PV are medians since S is centroid.
PQ = PU + UQ
PQ = 10
V is midpoint of RQ so RV = VQ, VQ = 5
RQ = RV + VQ
RQ = 10
T is midpoint of PR so RT = TP, TP = 3
RP = RT + TP
RP = 6
Perimeter = PQ + QR + RP =
= 26

27. Use of Altitudes
Altitude of a triangle – perpendicular segment from a vertex to the opposite side or line that contains the opposite side (may have to extend the side of the triangle)
Orthocenter – point at which the lines containing the three altitudes of a triangle intersect
Acute = inside of
Right = On the
Obtuse = Outside of

28. Altitude - Special Segment of Triangle
The perpendicular segment from a vertex of the triangle to the segment that contains the opposite side.
Definition: B A D F
In a right triangle, two of the altitudes are the legs of the triangle. B A D F I K
In an obtuse triangle, two of the altitudes are outside of the triangle.

29. Concurrency of Altitudes of a Triangle (Orthocenter)
The lines containing the altitudes of a triangle are concurrent.
Two or more lines are said to be concurrent if they intersect in a single point.
Every triangle has 3 altitudes that will always intersect in the same point.
Notice obtuse triangle, orthocenter is outside the triangle.
Notice right triangle, orthocenter is on the triangle.

30. Altitudes of Triangles
If the triangle is acute, then the altitudes are all in the interior of the triangle.

31. Altitudes of Triangles
If the triangle is a right triangle, then one altitude is in the interior and the other 2 altitudes are the legs of the triangle.

32. Altitudes of Triangles
If the triangle is an obtuse triangle, then one altitude is in the interior and the other 2 altitudes are in the exterior of the triangle.

33. Use Angle Bisectors of Triangles
Angle bisector – a ray that divides an angle into two congruent adjacent angles
Angle Bisector of a Triangle: A segment is an angle bisector of a triangle if one endpoint is a vertex of the triangle and the other endpoint is any other point on the triangle such that the segment bisects an angle of the triangle.
Incenter – point of concurrency of the three angle bisectors
**NOTE: In geometry, distance means the shortest length between two objects and this is always perpendicular. **

34. ANGLE BISECTOR THEOREM
If a point is on the bisector of an angle, then it is equidistant from the two sides of the angle.
If AD bisects BAC and DBAB and DCAC, then DB = DC
CONVERSE OF THE ANGLE BISECTOR THEOREM
If a point is in the interior of an angle and is equidistant from the sides of the angle, it lies on the angle bisector of the angle.
If DBAB and DCAC and DB = DC, then AD is the
of BAC.
angle bisector

35. Angle Bisector Theorem
If D is on the bisector of ∠ABC, then X D B Y C
DX = DY.

36. Angle Bisector Converse TheoremX
If WX = WZ, then W is on the bisector of ∠XYZ. W Y Z

37. Example 1 Use the Angle Bisector Thm Find the length of LM.
JM bisects < KJL because m < K JM = m < L JM.
ML = MK
ML = 5
Because JM bisects < KJL and MK  JK and ML  JL
Substitution
Example 2 Use Algebra to solve a problem
For what value of x does P lie on the bisector of < GFH?
P lies on the bisector of < GFH if m < GFP = m < HFP.
m < GFP = m < HFP
13x = 11x + 8
x = 4
Set angle measures equal.
Substitute.
Solve for x.

38. Angle Bisectors of Triangles
Every triangle has 3 angle bisectors which will always intersect in the same point - the incenter. The incenter is the same distance from all 3 sides of the triangle. The incenter of a triangle is also the center of a circle that will intersect each side of the triangle in exactly one point.

39. Angle Bisectors of Triangles

40. Angle Bisectors of Triangles

41. Angle Bisectors of Triangles

42. 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. If AP, BP, and CP are angle bisectors of ∆ ABC, then PD =
The point of concurrency is called the incenter.
Note: The incenter is always “inside” of the triangle.
Note: The incenter is equal distance from all three sides.
PE = PF

43. VS = VT = VU
a2 + b2 = c2
152 + VT2 = 172
225 + VT2 = 289
VT2 = 64
VT = 8
VS = 8
Theorem
Pythagorean Theorem
Substitute known values.
Multiply.
Subtract 225 from both sides.
Take Square Root of both sides.
Substitute.

44. equidistant LI c2
a2 + b2
225 = LI
152 LI
– 144 = – 144 81
LI2
81 = LI2 9 LI LI 9

45. In the diagram, D is the incenter of ∆ABC. Find DF.
Solve for x.
Solve for x.
Because angles are congruent and the segments are perpendicular, then the segments are congruent.
10 = x + 3
x = 7
Because segments are congruent and perpendicular, then the angle is bisected which means they are are congruent.
9x – 1 = 6x + 14
3x = 15
x = 3
In the diagram, D is the incenter of ∆ABC. Find DF.
DE = DF = DG
DF = DG
DF = 3
Concurrency of Angle Bisectors
Substitution

46. In ΔABC below, AB ≅ BC and AD bisects ∠BAC
In ΔABC below, AB ≅ BC and AD bisects ∠BAC. If the length of BD is 3(x + 2) units and BC = 42 units, what is the value of x? 5 6 12 13 A C D B

47. Perpendicular Bisector of a Triangle
A line or line segment that passes through the midpoint of a side of a triangle and is perpendicular to that side.
Perpendicular Bisector

48. Median of a Triangle
A segment that joins a vertex of the triangle and the midpoint of the opposite side.
Median

49. Altitude of a Triangle
A segment from a vertex of the triangle to the line containing the opposite side and perpendicular to the line containing that side.
Altitude

50. Angle Bisector of a Triangle
A segment that bisects an angle of the triangle and has one endpoint at a vertex of the triangle and the other endpoint at another point on the triangle.
Angle Bisector

51. Example Name all segments that are (if any) Angle Bisectors
Perpendicular Bisectors
Altitudes
Medians