Essential Question – What special segments exist in triangles?

5-2 Use Perpendicular BisectorsPerpendicular bisector – a segment, ray,

line, or plane that is perpendicular to a segment at its midpoint

Equidistant - same distanceCircumcenter – the point of concurrency

of the 3 perpendicular bisectors of a triangle

TBK p.264

THEOREM 5.2: Perpendicular Bisector TheoremIf 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 TheoremIf 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

TBK p.264

JM = ML Bisector Thm 7x = 3x + 16

– 3x = – 3x4x = 16 4 = 4

SubstituteSolve for x

x = 4ML = 3x + 16 = 3(4) + 16 = 12 + 16 = 28

AC = CBAE = BEAD = BD

DC bisects ABDefinition 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.

TBK p.265

KG = KH, JG = JH, FG = FH

KG = KH2x = x + 1-x -xx = 1

GH = KG + KHGH = 2x + (x + 1)GH = 2(1)+ (1 + 1)GH = 2+ (2)GH = 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.

Note: The circumcenter can be inside or outside of the triangle

Note: The circumcenter is equal distance from all the vertices.

CC

5.3 Use Angle Bisectors of TrianglesAngle bisector – a ray that divides an angle into two

congruent adjacent anglesIncenter – point of concurrency of the three angle

bisectors

**NOTE: In geometry, distance means the shortest length between two objects and this is always perpendicular. **

NTG p.282

THEOREM 5.5: ANGLE BISECTOR THEOREMIf 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 DBAB and DCAC, then DB = DC

THEOREM 5.5: CONVERSE OF THE ANGLE BISECTOR THEOREMIf 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 DBAB and DCAC and DB = DC, then AD is the of BAC.

angle bisector

Example 1 Use the Angle Bisector Thm

Find the length of LM.

JM bisects KJL because m K JM = m L JM.ML = KLML = 5

Because JM bisects KJL and MK JK and ML JLSubstitution

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 mGFP = mHFP.

mGFP = mHFP

13x = 11x + 8

x = 4

Set angle measures equal.

Substitute.Solve for x.

Do the angle bisector part of the taskThen 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 =

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

VS = VT = VU

a2 + b2 = c2

152 + VT2 + 172

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.

NTG p.282

equidistant

LI

c2 a2 + b2

LI2 + 122152 225 = LI2 + 144 – 144 = – 144

81 = LI281 LI2

9 LI

LI 9

Solve for x. Solve for x.

In the diagram, D is the incenter of ∆ABC. Find DF.

Because angles are congruent and the segments are perpendicular, then the segments are congruent.10 = x + 3x = 7

Because segments are congruent and perpendicular, then the angle is bisected which means they are are congruent.9x – 1 = 6x + 143x = 15x = 3

DE = DF = DGDF = DGDF = 3

Concurrency of Angle Bisectors

Substitution

AssignmentTextbook: p266 (1-18) & p274 (1-12)

5.4 Use Medians and AltitudesMedian of a triangle – a segment from a vertex

to the midpoint of the opposite sideCentroid – 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

TBK p.282

Acute = inside of

Obtuse = Outside of

Right = On the

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? 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

Now assume, A is vertex, C is centroid, and B is midpoint of opposite side. CB = 3

Vertex to centroid = 2/3 medianCentroid to midpoint = 1/3 median

P is the centroid of ABC.

The dist. from the vertex to the centroid is twice the dist. from centroid to midpoint. AP = 2 PE CP = 2 PDBP = 2

PF

PE = ½ AP PD = ½ CPPF = ½ BP

AP = 2/3 AE

CP = 2/3 CD

BP = 2/3 BF

PE = 1/3 AE

PD = 1/3 CD

PF = 1/3 BF

The dist. from the centroid to midpoint is half the dist. from the vertex to the centroid.The dist. from the vertex to the centroid is 2/3 the distance of the median.The dist. from the centroid to midpoint is 1/3 the distance of the median.

What relationships exist?

What do I know about DG?

BGDG3

1BDDG

2

1

)12(2

1DG

DG = 6

What do I know about BG?

BGBD3

2 BG = BD +

DGBG = 12 + 6BG = 18

Your TurnIn PQR, S is the centroid, UQ = 5, TR = 3, RV = 5, and SU = 2.

1. Find RU and RS.

2. Find the perimeter of PQR.

RU is a median and RU = RS + SU.

RS = 2 SURS = 2 2RS = 4

RU = RS + SURU = 4 + 2RU = 6

Perimeter means add up the sides of the triangle.

T is midpoint of PR so RT = TP, TP = 3

U is midpoint of PQ so PU = UQ , PU = 5

V is midpoint of RQ so RV = VQ, VQ = 5

= 10 + 10 +6

QT, RU, and PV are medians since S is centroid.

RP = RT + TP

PQ = PU + UQRQ = RV + VQ

RP = 6

PQ = 10RQ = 10

Perimeter = PQ + QR + RP = 26

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.

AssignmentTextbook: p280-281 (1-6, 10-24)