1

MEDIANS AREPROPORTIONAL

ALTITUDES AREPROPORTIONAL

ANGLE BISECTORSARE PROPORTIONAL

PERIMETERS AREPROPORTIONAL

ANGLE BISECTORSFORM PROPOR. SEG.

PROBLEM 6

PROBLEM 7

PROBLEM 8

PROBLEM 9

STANDARDS 4 and 5

END SHOW

SPECIAL SEGMENTS

IN A TRIANGLE

PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

2

Standard 4:

Students prove basic theorems involving congruence and similarity.

Los estudiantes prueban teoremas básicos que involucran congruencia y semejanza.

Standard 5:

Students prove triangles are congruent or similar and are able to use the concept of corresponding parts of congruent triangles.

Los estudiantes prueban que triángulos son congruentes o semejantes y son capaces de usar el concepto de partes correspondientes de triángulos congruentes.

PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

3STANDARDS 4 and 5

ALTITUDE:

A segment from a vertex of a triangle perpendicular to the line containing the opposite side.

ALTURA:

Un segmento desde el vértice de un triángulo perpendicular a la línea conteniendo el lado opuesto.

PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

4STANDARDS 4 and 5

PERPENDICULAR BISECTOR:

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

BISECTRIZ PERPENDICULAR:

Una línea o segmento que pasa a través del punto medio de un lado de un triángulo y es perpendicular a ese lado.

PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

5STANDARDS 4 and 5

MEDIAN:

A segment that connects a vertex of a triangle to the midpoint of the opposite side.

MEDIANA:

Un segmento que conecta un vertice de un triángulo a el punto medio del lado opuesto.

PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

6STANDARDS 4 and 5

ANGLE BISECTOR:

A segment from a vertex to the opposite side that bisects (divides in two equal parts) the angle of the triangle.

BISECTRIZ ANGULAR:

Un segmento desde un vértice a el lado opuesto que biseca (divide en dos partes iguales) el ángulo de el triángulo.

PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

7STANDARDS 4 and 5

Draw an isosceles triangle and draw an altitude, a perpendicular bisector, a median and an angle bisector from the vertex angle. What do you observe?

Altitude

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8STANDARDS 4 and 5

Draw an isosceles triangle and draw an altitude, a perpendicular bisector, a median and an angle bisector from the vertex angle. What do you observe?

Perpendicualr Bisector

PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

9STANDARDS 4 and 5

Draw an isosceles triangle and draw an altitude, a perpendicular bisector, a median and an angle bisector from the vertex angle. What do you observe?

Median

PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

10STANDARDS 4 and 5

Draw an isosceles triangle and draw an altitude, a perpendicular bisector, a median and an angle bisector from the vertex angle. What do you observe?

Angle Bisector

PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

11STANDARDS 4 and 5

Draw an isosceles triangle and draw an altitude, a perpendicular bisector, a median and an angle bisector from the vertex angle. What do you observe?

Altitude

Perpendicualr Bisector

MedianAngle Bisector

Lets put them all together

So, all of them occupy the same Geometric Space in the triangle!

Could you do something similar with an equilateral triangle in all the vertices?

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12

A

BC

K

L

M

KLAB

MK

CALMBC

STANDARDS 4 and 5

CAB MKL

KLAB

MKCA

LMBC

OR

REVIEW:

PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

13

A

BC

K

L

M

ABKL

CA

MKBCLM

N

KNDAD

In simlar triangles MEDIANS are proportional to sides:

STANDARDS 4 and 5

KLAB

MKCA

LMBC

KNAD

OR

PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

14

J

I KS

J

UT R

TU

IK

STANDARDS 4 and 5

=JS

6 10

30=

IJK TJU, R is the midpoint of TU and S is the midpoint of IK. TU= 10, JR= 6, IK = 30. Find JS.

S

R

J

UT

I K

JS

JR

(6) (6)

=(30)(6)

10JS

= 180

10JS

JS=18

PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

15

A

BC

K

L

M

In similar triangles ALTITUDES are proportional to sides:

ABKL

CA

MK

BCLM

O

KOEAE

STANDARDS 4 and 5

KLAB

MK

CA

LMBC

KOAE

OR

PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

16

FAF

P

KP

A

BC

K

L

M

In similar triangles ANGLE BISECTORS are proportional to sides:

ABKL

CA

MK

BC

LM

STANDARDS 4 and 5

KLAB

MK

CA

LMBC

KPAF

OR

PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

17

JR

JS

JT

JT+TI

STANDARDS 4 and 5

S

R

J

UT

I K

J

I KS

J

UT R

JS is an angle bisector. If TJ = 14, IT=22, and JS= 30, what is the value for JR? Suppose IJK TJU.

14 36

22

14

= JR

30 36

14=

(30) (30)

=(14)(30)

36JR

= 420

36JR

30

JR = 11.7.

PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

18

A

BC

K

L

MABKL

CA

MK

BCLM

In similar triangles PERIMETERS proportional to sides:

KL MK LM+ +AB CA BC+ +

STANDARDS 4 and 5

KLAB

MK

CA

LMBC

KL MK LM+ +AB CA BC+ +

OR

PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

19

JK

JU JT+JU+TU

PERIMETER IJK

STANDARDS 4 and 5

S

R

J

UT

I K

Given: JT=20, JU=30, TU=70, AND JK=90, what is the perimeter of IJK? Suppose TJU IJK.

J

I KS

J

UT R

=

90

30 20+30+70

PERIMETER IJK=

120

PERIMETER IJK=3(120) (120)

PERIMETER IJK = 360

20 30

7090

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20

A

B CD

ABACDC

BD

STANDARDS 4 and 5PRESENTATION CREATED BY SIMON PEREZ. All rights reserved

21

IS

SK

JI

JK

STANDARDS 4 and 5

S

R

J

UT

I K

JS is an angle bisector. JI= X+5, JK=X+3, IS=3, and SK=2. Find the value for X. Suppose IJK TJU.

= =

3(X+3) = 2(X+5)

3X + 9 = 2X + 10-9 -9

3X = 2X + 1-2X -2X

X = 1

X+5

X+32

3

X+3X+5

23

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