Download GEOMETRY JOURNAL. 5. christa walters

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TRIANGLES…….
…
TRIANGLES….
…
AND…
…
MORE…
TRIANGLES.
Menu options ahead.
Option
#1
What is it?
A line that bisects a segment and is perpendicular
to that same segment.
THEOREM:
* Any point that is on the perpendicular bisector
is equidistant to both of the endpoints of the
segment which is perpendicular to.
Converse:
*If the point is equidistant from both of the endpoints of
the segment, then it is a perpendicular bisector.
Extra:
A perpendicular bisector is the locus of all points in a plane that are equidistant
from the endpoints of a segment.
Segment
Perpendicular
bisector
Are both
congruent.
1. Given that AC= 25.5, BD=20,
and BC= 25.5. Find AB.
a
AB= 51
2. Given that m is the
perpendicular bisector of AB and
BC= 30 . Find AC.
m
d
c
AC= 30
3.
b
Given that m is the
perpendicular bisector of AB ,
AC= 4a, and BC= 2a+26. Find
BC.
2a+26=4a
-2a=-26
a=13 plug in both equations
BC= 52
Option
#2
What is it?
A ray or any line that cuts an angle into two
congruent angles.
Theorem :
* Any point that lies on the angle bisector is
equidistant to both of the sides of the angle.
Converse:
* If it is equidistant to both sides of the angle, then it
lies on the angle bisector.
MUST: the point always has to lie on the
interior of the angle.
Angle
point that
lies on the bisector
is equidistant to both
sides of the angle
Angle bisector
1. Given that BD bisects ∠ABC and
CD = 21, find AD.
AD= 21
A
2. Given that AD =65 , CD= 65 and
m∠ABC = 50˚, find m∠CBD.
D
m∠DBC= 25˚
3. Given that DA=DC , m∠DBC=
(10y+3)˚ , and m ∠DBA=
(8y+10) ˚, find m∠DBC.
B
C
10y+3=8y+10
2y+3=10
2y=7 y=3.5 plug in both
equations.
m∠DBC=38˚
Option
#3
example. G is the point of concurrency.
It is the point where all the 3 perpendicular bisectors of a triangle meet.
What makes it special?
It is the point of concurrency and it is equidistant to all 3 vertices.
1. Knowing what circumcenter is, it would help you if you are a
business person to know a strategic place in between 3 cities to
place your business. That way it would be located in the same
distance from all 3 cities for customers to come.
EXAMPLE
F is the circumcenter in this
equilateral triangle. The
circumcenter is on the inside of
the triangle.
2.In an acute the circumcenter is on
the inside of the triangle
C is the circumcenter
3.In a right the circumcenter is on
the midpoint of the hypotenuse
D is the circumcenter
4.In an obtuse the circumcenter is
on the outside of the triangle.
F is the circumcenter
c
d
f
The angle bisectors of a triangle meet at a
point that is equidistant from the three sides of
the triangle.
1. What is the distance from H to G,
if F to H is 10 cm?
10cm from H to G
2. H to E is 15cm , then what would
be H to G doubled?
Real: H to G would be 15cm and
doubled would be 30cm
3. H in this triangle is what?
The incenter of triangle ABC
TRIANGLE
•The point of concurrency where the three angle bisectors of a triangle
meet. It is equidistant (same distance) from each three sides of the
triangle
EXAMPLES:
2. If you want to open a new
restaurant in town you have
to place it at the incenter of
three major highways so
people can go to your
restaurant.
G.
M
K
P
J
L
1. P is the incenter
H
3. If the distance from K
to P is 2cm, what is the
distance from P to L?
From P to L = 2cm
Option
#4
Ex.1 :
Median
Is the segment that goes from the vertex of
a triangle to the opposite midpoint
Ex. 2: all medians are
congruent. Shown in yellow
Ex.3 :
Median
The point where the medians of a triangle
intersect.
Is the center of balance. The distance from the vertex to the
centroid is the double the distance from the centroid to the
opposite side.
EXAMPLES:
1. Balance a triangular piece of
glass on a triangular coffee table.
2. G is the centroid.
3.
AP= 2/3 AY
BP=2/3BZ
CP=2/3CX
b
y
x
p
c
z
a
The medians of a triangle intersect at a point that is 2/3 of the
distance from each vertex to the midpoint of the opposite side
Option #5
A segment that goes from the vertex perpendicular
to the line containing the opposite side.
Where the altitudes of a triangle intersect.
Concurrency
of Altitudes of
a Triangle
Theorem
The lines
containing the
altitudes of a
triangle are
concurrent
Option
#6
A segment that joins the midpoints
of two sides of the triangle.
Every triangle has 3midsegments
that form the midsegment triangle.
Midsegment Theorem The segment joining the midpoints of two sides
of a triangle is parallel to the third side, and is half the length of the
third side.
Option #7
EXAMPLES:
In any triangle the longest side is always opposite from
the largest angle. The small side is opposite from a
small angle.
Option #8
The exterior angle is
bigger (grater) than the
non-adjacent interior
angles of the triangle .
TELL WHEATHER A TRIANGLE CAN HAVE THE SIDES WITH THE GIVEN LENGTHS. EXPLAIN
1. 3,5,7
3+5>7
8>7
2. 4, 6.5,11
4 +6.5> 11
10.5>11
YES, the sum of each
Pair of the lengths is
Grater than the 3rd <.
NO. by the 3rd triangle
inequality thm. A triangle
Cannot have these side lengths.
3. 7, 4, 5
5+4 > 7
9>7
Yes , 5+4 is grater than 7 so
there can be a triangle.
Triangle Inequality Theorem The sum of
the lengths of two sides of a triangle is
greater than the length of the third side.
Option #9
1. Assume that what you are proving is FALSE
2. Use that as your given, and start proving it.
3. When you come to a contradiction that’s when you stop and have proven that it is true.
PROVE: A TRIANGLE
CANNOT HAVE 2 RIGHT
ANGLES
PROVE: A SCALENE
TRIANGLE CANNOT HAVE
TWO CONGRUENT
ANGLES
1. assume a triangle has
1. Assume a scalene
2 right angles <1+<2….
triangle can have two
GIVEN
congruent sides
2. M<1 =M<2 = 90.. DEFF
2. a triangle with 2
OF RT. <
congruent angles…
3. M<1+M<2+M<3=180
deff. of an isosceles
…..TRIANGLE + THM
triangle
4. M<3= 0
3. Deff. Of isosceles and
CONTRADICTION
scalene triangles…
therefore a scalene
So a triangle cannot have
triangle cannot have
2 right angles.
two congruent angles.
PROVE: IF A>0
THAN 1/A >0
1. Assume a>0 than
1/a<0
2. (a) 1/a <0a
a 1 <0
CONTRADICTION
3. Therefore if a>0
than 1/a>0
Option #19
If 2 triangles have 2 sides that are congruent, but the third side is not congruent,
then the triangle with the large included angle has the longer side.
(What changes is the incidence and the side….)
Theorem:
If two sides of one triangle are congruent to the two sides of another triangle and
the third sides are not congruent, then the larger included angle is across from the
longer third side.
Converse :
Option #11
45
60
Both legs of the triangle
are congruent and the
length of the hypotenuse
is the length of the two
legs √2
90
45
The length of the
hypotenuse is 2x the
length of the shorter
leg, and the length of
the longer leg is the
length of the shorter
leg time √3
18√3
y
x
60
y
90
1. 18√3 =2x
9 √3=x
y=x √3
y=27
60
x
30
2 . 15= √3
15/ √3=x
5 √3=x
y= 2x
Y=2(5 √3)
Y= 10 √3
30
X=5 √3
Y=2(5)
Y=10
5
30
y
"The only angle from which to approach a
problem is the TRY-Angle"