Transcript Points of Concurrency in Triangles (5

Essential Question –
What special segments exist in triangles?
TBK p.264
5-2 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
 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 perp bisector CP.
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 = 12 + 16 = 28
TBK p.265
AC = CB
AE = BE
AD = BD
DC bisects AB
Definition of Congruence
Bisector Thm.
Yes, because AE = BE, E is equidistant from B and A. According
to Converse of Bisector, E is on the Bisector DC.
KG = KH, JG = JH, FG = FH
KG = KH
2x = x + 1
-x -x
x =1
GH
GH
GH
GH
GH
=
=
=
=
=
KG + KH
2x + (x + 1)
2(1)+ (1 + 1)
2+ (2)
4
 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.
The perpendicular bisectors are
concurrent at a point called the
circumcenter.
CC
Note: The circumcenter can be
inside or outside of the triangle
Note: The circumcenter is equal
distance from all the vertices.
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. **
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
angle bisector of BAC.
Example 1 Use the Angle Bisector Thm
Find the length of LM.
JM bisects KJL because m K JM = m L JM.
Because JM bisects KJL and MK
ML = KL
Substitution
ML = 5
JK and ML
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 Set angle measures equal.
13x = 11x + 8
x=4
Substitute.
Solve for x.
JL
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.
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 = PE = PF
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.
VS = VT = VU
Thm 5.7
a2 + b2 = c2
Pythagorean Thm.
152 + VT2 + 172 Substitute known values.
225 + VT2 = 289 Multiply.
VT2 = 64
Subtract 225 from both sides.
VT = 8
Take Square Root of both sides.
VS = 8
Substitute.
NTG p.282
equidistant
LI
c2
152
81
9
a2 + b2
LI2
+
225 = LI2 + 144
– 144 =
– 144
81 = LI2
122
LI2
LI
LI
9
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
Assignment
 Textbook: p266 (1-18) & p274 (1-12)
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
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.
Why?
Now assume, A is vertex, C is
centroid, and B is midpoint of
opposite side.
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 Vertex to centroid = 2/3 median
CB = 3 Centroid to midpoint = 1/3 median
What do you notice about AC and CB?
AC is twice CB
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
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
The dist. from the centroid
to midpoint is 1/3 the
distance of the median.
PE = 1/3
AE
BP = 2/3
BF
PF = 1/3
BF
CP = 2/3
CD
PD = 1/3
CD
PD
What do I know about DG?
1
1
DG  BD
DG  BG
2
3
1
DG  (12)
2
DG = 6
What do I know about BG?
2
BD  BG
3
BG = 12 + 6
BG = 18
BG = BD + DG
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
RS = 2
RS = 4
SU
2
RU = RS + SU
RU = 4 + 2
RU = 6
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
RQ = RV + VQ
V is midpoint of RQ so RV = VQ, VQ = 5
T is midpoint of PR so RT = TP, TP = 3
RP = RT + TP
Perimeter = PQ + QR + RP = 10 + 10 +6 = 26
PQ = 10
RQ = 10
RP = 6
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 right
triangle,
orthocenter is on
the triangle.
Notice obtuse
triangle,
orthocenter is
outside the
triangle.
Assignment
 Textbook: p280-281 (1-6, 10-24)