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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
PRESENTATION CREATED BY SIMON PEREZ. All rights reserved
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?
PRESENTATION CREATED BY SIMON PEREZ. All rights reserved
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
PRESENTATION CREATED BY SIMON PEREZ. All rights reserved
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
PRESENTATION CREATED BY SIMON PEREZ. All rights reserved