1. Points of Concurrency in Triangles (5-1 through 5 – 4)
Essential Question –
What special segments exist in triangles?

2. 5-2 Use Perpendicular Bisectors
TBK p.264
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. THEOREM 5.2: 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
THEOREM 5.3: 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. TBK p.264
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

5. TBK p.265 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.

6. 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

7. Do the perpendicular bisector part of the task
Then show the geosketch example, showing how the center can move depending on the type of triangle used.

8. The perpendicular bisectors are concurrent at a point called the circumcenter.
Note: The circumcenter can be inside or outside of the triangle
Note: The circumcenter is equal distance from all the vertices.

9. 5.3 Use Angle Bisectors of Triangles
NTG p.282
5.3 Use Angle Bisectors of Triangles
Angle bisector – a ray that divides an angle into two congruent adjacent angles
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. **

10. THEOREM 5.5: 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
THEOREM 5.5: 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

11. Example 1 Use the Angle Bisector Thm Find the length of LM.
JM bisects KJL because m K JM = m L JM.
ML = KL
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.

12. Do the angle bisector part of the task
Then show the geosketch example, showing how the center can move depending on the type of triangle used.

13. THEOREM 5.7: 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

14. VS = VT = VU
a2 + b2 = c2
152 + VT
225 + VT2 = 289
VT2 = 64
VT = 8
VS = 8
Thm 5.7
Pythagorean Thm.
Substitute known values.
Multiply.
Subtract 225 from both sides.
Take Square Root of both sides.
Substitute.

15. equidistant LI c2 a2 + b2 225 = LI2 + 144 152 LI2 + 122 – 144 = – 144
NTG p.282
equidistant LI c2
a2 + b2
225 = LI
152 LI
– 144 = – 144 81
LI2
81 = LI2 9 LI LI 9

16. 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

17. Assignment
Textbook: p266 (1-18) & p274 (1-12)

18. 5.4 Use Medians and Altitudes
TBK p.282
5.4 Use Medians and Altitudes
Median of a triangle – a segment from a vertex to the midpoint of the opposite side
Centroid – point of concurrency of the three medians of a triangle (always inside the )
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 

19. Theorem 5.8 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

20. P is the centroid of  ABC.
What relationships exist?
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

21. 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

22. 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

23. Theorem 5.9 Concurrency of Altitudes of a Triangle (Orthocenter)
TBK p.278
Theorem 5.9 Concurrency of Altitudes of a Triangle (Orthocenter)
The lines containing the altitudes of a triangle are concurrent.
In a right triangle, the legs are also altitudes.
In an obtuse triangle, sides of the triangle and/or the altitudes may have to be extended.
Notice obtuse triangle, orthocenter is outside the triangle.
Notice right triangle, orthocenter is on the triangle.

24. Assignment
Textbook: p (1-6, 10-24)