Points of Concurrency in Triangles (5-1 through 5

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Points of Concurrency in Triangles(5-1 through
5 4)

• Essential Question
• What special segments exist in triangles?

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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

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

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• 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.

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5.3 Use Angle Bisectors of Triangles
NTG p.282

• 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.

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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
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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.
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• 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.

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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
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Thm 5.7 Pythagorean Thm. Substitute known
values. Multiply. Subtract 225 from both
sides. Take Square Root of both sides. Substitute.
VS VT VU a2 b2 c2 152 VT2 172 225
VT2 289 VT2 64 VT 8 VS 8
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NTG p.282
equidistant
LI
a2 b2
c2
225 LI2 144
LI2 122
152
144 144
LI2
81
81 LI2
LI
9
LI
9
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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
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Assignment

• Textbook p266 (1-18) p274 (1-12)

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5.4 Use Medians and Altitudes
TBK p.282

• 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 ?
Obtuse ? Outside of ?
Right ? On the ?
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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
Centroid to midpoint 1/3 median
CB 3
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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
CP 2 ? PD
BP 2 ? PF
The dist. from the centroid to midpoint is half
the dist. from the vertex to the centroid.
PE ½ ? AP
PD ½ ? CP
PF ½ ? BP
The dist. from the vertex to the centroid is 2/3
the distance of the median.
AP 2/3 ? AE
CP 2/3 ? CD
BP 2/3 ? BF
The dist. from the centroid to midpoint is 1/3
the distance of the median.
PE 1/3 ? AE
PD 1/3 ? CD
PF 1/3 ? BF
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What do I know about DG?
What do I know about BG?
BG BD DG
BG 12 6
BG 18
DG 6
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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
RQ RV VQ
RQ 10
V is midpoint of RQ so RV VQ, VQ 5
T is midpoint of PR so RT TP, TP 3
RP RT TP
RP 6
10 10 6
Perimeter PQ QR RP
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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.
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Assignment

• Textbook p280-281 (1-6, 10-24)