Transcript 5.3 Using Angle Bisectors of Triangles

5.3 Using Angle Bisectors
of Triangles
Vocabulary/Theorems
 Angle bisector: ray that divides angle
into 2 congruent angles
 Point of concurrency: point of
intersection of segments, lines, or rays
 Incenter: point of concurrency of angle
bisectors of a triangle
 Angle Bisector Theorem: If a point is on
the bisector of an angle, then it is
equidistant from the 2 sides. (distance
from point to a line is a perp. path)
Vocabulary/Theorems
Converse of Angle Bisector
Theorem:
Angle bisectors intersect at a point
that is equidistant from the sides of
a triangle. (Incenter is equidistant
from sides)
To Construct Angle Bisectors
 Place point of compass on the angle
vertex
 Draw an arc through both adjacent sides
of the triangle
 Move the point of the compass to one of
the intersection points of the arc and the
side
 Open the compass ½ the distance
between the 2 sides and create an arc
above the current one
 Move the compass to the other side and
repeat making the 2 arcs intersect
 Using a straightedge, connect the vertex
with this new arc intersection
To Construct Angle Bisectors
If repeated with the 3
angle, the bisectors
would meet at the
point of concurrency,
the incenter.
The incenter is
equidistant to each
side of the triangle.
EXAMPLE 1
Use the Angle Bisector Theorems
Find the measure of GFJ.
SOLUTION
Because JG FG and JH FH and
JG = JH = 7, FJ bisects GFH by the
Converse of the Angle Bisector Theorem.
So,
mGFJ = mHFJ = 42°.
EXAMPLE 2
Solve a real-world problem
A soccer goalie’s position relative to the ball and
goalposts forms congruent angles, as shown. Will the
goalie have to move farther to block a shot toward
the right goalpost R or the left goalpost L?
SOLUTION
The congruent angles tell you that the goalie is on
the bisector of LBR. By the Angle Bisector
Theorem, the goalie is equidistant from BR and BL .
So, the goalie must move the same distance to
block either shot.
With a partner, do #1-3 on p. 273
EXAMPLE 3
Use algebra to solve a problem
For what value of x does P lie on the
bisector of A?
SOLUTION
From the Converse of the Angle
Bisector Theorem, you know that P lies
on the bisector of A if P is equidistant
from the sides of A, so when BP = CP.
BP = CP
x + 3 = 2x –1
4=x
Set segment lengths equal.
Substitute expressions for segment lengths.
Solve for x.
Point P lies on the bisector of
A when x = 4.
for Examples 1, 2, and 3
GUIDED PRACTICE
In Exercises 1–3, find the value of x.
1.
2.
B
P
A
C
ANSWER
15
B
P
A
C
ANSWER
11
P
3.
B
C
A
ANSWER
5
GUIDED PRACTICE
for Examples 1, 2, and 3
4. Do you have enough information to conclude that
QS bisects
PQR? Explain.
ANSWER
No; you need to establish that SR
SP
QP.
QR and
Do #5 on p. 273
EXAMPLE 4
Use the concurrency of angle bisectors
In the diagram, N is the incenter of
ABC. Find ND.
SOLUTION
By the Concurrency of Angle
Bisectors of a Triangle Theorem, the
incenter N is equidistant from the
sides of ABC. So, to find ND, you
can find NF in NAF. Use the
Pythagorean Theorem stated on page
18.
EXAMPLE 4
Use the concurrency of angle bisectors
c 2 = a 2 + b2
2
2
20 = NF + 16
2
Pythagorean Theorem
2
400 = NF + 256
144 = NF
12 = NF
2
Substitute known values.
Multiply.
Subtract 256 from each side.
Take the positive square root of each side.
Because NF = ND, ND = 12.