Section 5.1Bisectors of Triangles

We learned earlier that a segment bisector is any line, segment, or plane that intersects a segment at its midpoint. If a bisector is also perpendicular to the segment, it is called a perpendicular bisector.

Example 1:

a) Find the length of BC.

From the information in the diagram,

we know that is the perpendicular

bisector of .

CD

AB

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BC = AC Perpendicular Bisector TheoremBC = 8.5 Substitution

Example 1:

b) Find the length of XY.

Since and , is the

perpendicular bisector of by the converse

of the Perpendicular Bisector Theorem. By

the definition of segment bisector, .

Since 6, 6.

WX WZ WY XZ WY

XZ

XY ZY

ZY XY

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Example 1:

c) Find the length of PQ.

is the perpendicular bisector of .SQ PR�������������� �

PQ = RQ Perpendicular Bisector Theorem3x + 1 = 5x – 3 Substitution1 = 2x – 3 Subtract 3x from each side.4 = 2x Add 3 to each side.2 = x Divide each side by 2.

So, PQ = 3(2) + 1 = 7

When three or more lines intersect at a common point, the lines are called concurrent lines. The point where concurrent lines intersect is called the point of concurrency. A triangle has three sides, so it also has three perpendicular bisectors. These bisectors are concurrent lines. The point of concurrency of the perpendicular bisectors is called the circumcenter of the triangle.

The circumcenter can be on the interior, exterior, or side of a triangle.

Example 2: GARDEN A triangular-shaped garden is shown. Can a fountain be placed at the circumcenter and still be inside the garden?

By the Circumcenter Theorem, a point equidistant from three points is found by using the perpendicular bisectors of the triangle formed by those points.

No, the circumcenter of an obtuse triangle is in the exterior of thetriangle.

Copy ΔXYZ, and use a ruler and protractor to draw the perpendicular bisectors. The location for the fountain is C, the circumcenter of ΔXYZ, which lies in the exterior of the triangle.

C

We learned earlier that an angle bisector divides an angle into two congruent angles. The angle bisector can be a line, segment, or ray.

Example 3:

a)Find the length of DB.

DB = DC Angle Bisector Theorem

DB = 5 Substitution

Example 3:

b) Find mWYZ.

Since , , , is equidistant from the

sides of . By the Converse of the Angle Bisector Theorem,

bisects .

WX YX WZ YZ WX WZ W

XYZ

YW XYZ

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WYZ XYW Definition of angle bisectormWYZ = mXYW Definition of congruent anglesmWYZ = 28° Substitution

Example 3:

c) Find the length of QS.

QS = SR Angle Bisector Theorem

4x – 1 = 3x + 2 Substitution

x – 1 = 2 Subtract 3x from each side.

x = 3 Add 1 to each side.

So, QS = 4(3) – 1 or 11.

The angle bisectors of a triangle are concurrent, and their point of concurrency is called the incenter of a triangle.

Example 4:

a) Find ST if S is the incenter of ΔMNP.

By the Incenter Theorem, since S is equidistant from the sides of ΔMNP, ST = SU.

Find ST by using the Pythagorean Theorem.

a2 + b2 = c2 Pythagorean Theorem

82 + SU2 = 102 Substitution

64 + SU2 = 100 82 = 64, 102 = 100

Since length cannot be negative, use only the positive square root, 6. Since ST = SU, ST = 6.

SU2 = 36 Subtract 64 from each side.

SU = ±6 Take the square root of each side.

Example 4:

b) Find mSPU if S is the incenter of ΔMNP.

Since MS bisects RMT, mRMT = 2mRMS. So mRMT = 2(31) or 62°. Likewise, mTNU = 2mSNU, so mTNU = 2(28) or 56°.

mUPR + mRMT + mTNU = 180° Triangle Angle Sum Theorem

mUPR + 62° + 56° = 180° SubstitutionmUPR + 118° = 180° Simplify.

mUPR = 62° Subtract 118° from each side.

Since PS bisects UPR, 2mSPU = mUPR. This means that

mSPU = mUPR. __12

mSPU = (62) or 31°__12