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TEKS: G2A, G3B
The student will use constructions to explore
attributes of geometric figures and make
conjectures about relationships.
The student will construct and justify
statements about geometric figures and their
properties.
Note
Based on these theorems, an angle bisector can be
defined as the locus of all points in the interior of
the angle that are equidistant from the sides of the
angle.
Example: 1
Find the measure.
BC
BC = DC
 Bisector Thm.
BC = 7.2
Substitute 7.2 for DC.
Example: 2
Find the measure.
mEFH, given that mEFG = 50°.
Since EH = GH,
and
,
bisects
EFG by the Converse
of the Angle Bisector Theorem.
Def. of  bisector
Substitute 50° for mEFG.
Example: 3
Find mMKL.
Since, JM = LM,
and
bisects JKL
,
by the Converse of the Angle
Bisector Theorem.
mMKL = mJKM
Def. of  bisector
3a + 20 = 2a + 26
Substitute the given values.
a + 20 = 26
a=6
So mMKL = [2(6) + 26]° = 38°
Subtract 2a from both sides.
Subtract 20 from both sides.
Example: 4
Given that YW bisects XYZ and
WZ = 3.05, find WX.
WX = WZ
WX = 3.05
So WX = 3.05
 Bisector Thm.
Substitute 3.05 for WZ.
Example: 5
Given that mWYZ = 63°, XW = 5.7, and
ZW = 5.7, find mXYZ.
mWYZ + mWYX = mXYZ
mWYZ = mWYX
mWYZ + mWYZ = mXYZ
2mWYZ = mXYZ
2(63°) = mXYZ
126° = mXYZ
 Bisector Thm.
Substitute m WYZ for
mWYX .
Simplify.
Substitute 63° for mWYZ .
Simplfiy .
Example: 6
John wants to hang a spotlight along
the back of a display case. Wires AD
and CD are the same length, and A
and C are equidistant from B. How do
the wires keep the spotlight
centered?
It is given that
. So D is on the perpendicular bisector of
by the
Converse of the Angle Bisector Theorem. Since B is the midpoint of
,
is the perpendicular bisector of
. Therefore the spotlight remains
centered under the mounting.
Example: 7
S is equidistant from each pair of suspension lines.
What can you conclude about QS?
QS bisects  PQR.
A triangle has three angles, so it has three angle
bisectors. The angle bisectors of a triangle are also
concurrent. This point of concurrency is the
incenter of the triangle .
Remember!
The distance between a point and a line is the
length of the perpendicular segment from the
point to the line.
Unlike the circumcenter, the incenter is always inside
the triangle.
The incenter is the center of the triangle’s inscribed
circle. A circle inscribed in a polygon intersects each
line that contains a side of the polygon at exactly one
point.
Example: 8
MP and LP are angle bisectors of ∆LMN.
distance from P to MN.
Find the
P is the incenter of ∆LMN. By the Incenter Theorem, P is equidistant from
the sides of ∆LMN.
The distance from P to LM is 5. So the distance from P to MN is also 5.
Example: 8 cont.
MP and LP are angle bisectors of
∆LMN. Find mPMN.
mMLN = 2mPLN
PL is the bisector of MLN.
mMLN = 2(50°) = 100°
Substitute 50° for mPLN.
mMLN + mLNM + mLMN = 180°
100 + 20 + mLMN = 180
Δ Sum Thm.
Substitute the given values.
mLMN = 60°
Subtract 120° from both
sides.
PM is the bisector of LMN.
Substitute 60° for mLMN.
mPMN = 30 °
Example: 9
QX and RX are angle bisectors of ΔPQR. Find the
distance from X to PQ.
X is the incenter of ∆PQR. By the Incenter Theorem, X is equidistant from
the sides of ∆PQR.
The distance from X to PR is 19.2. So the distance from X to PQ is
also 19.2.
Example: 9 cont.
QX and RX are angle bisectors of
∆PQR. Find mPQX.
mQRY= 2mXRY
XR is the bisector of QRY.
mQRY= 2(12°) = 24°
Substitute 12° for mXRY.
mPQR + mQRP + mRPQ = 180°
mPQR + 24 + 52 = 180
mPQR = 104°
∆ Sum Thm.
Substitute the given values.
Subtract 76° from both
sides.
QX is the bisector of PQR.
Substitute 104° for mPQR.
Example: 10
A city plans to build a firefighters’ monument in the
park between three streets. Draw a sketch to show
where the city should place the monument so that it
is the same distance from all three streets. Justify
your sketch.
By the Incenter Thm., the incenter of
a ∆ is equidistant from the sides of
the ∆. Draw the ∆ formed by the
streets and draw the  bisectors to
find the incenter, point M. The city
should place the monument at point
M.