Math Core

Where should three towns build a shared water tower so it's the same distance from each? Where can you place the largest circular fountain inside a triangular plaza? Both questions are answered by bisectors: lines that cut a segment or an angle exactly in half. In this lesson you'll learn what makes the points on a bisector special, and where the bisectors of a triangle meet.

Perpendicular bisectors

A perpendicular bisector of a segment is a line (or ray or segment) that is perpendicular to the segment at its midpoint. Every point on it has a special property.

Line ℓ (dashed) is the perpendicular bisector of AB. Point P on ℓ is 5 units from both A and B.

Why is $PA = PB$ ? Triangles $PMA$ and $PMB$ share side $\overline{PM}$ , have right angles at $M$ , and $MA = MB$ because $M$ is the midpoint. So $\triangle PMA \cong \triangle PMB$ by SAS, and $\overline{PA} \cong \overline{PB}$ because corresponding parts of congruent triangles are congruent.

Perpendicular Bisector Theorem and its converse

Theorem: If a point is on the perpendicular bisector of a segment, then it is equidistant from the segment's endpoints.
Converse: If a point is equidistant from the endpoints of a segment, then it is on the segment's perpendicular bisector.

So the perpendicular bisector of $\overline{AB}$ is exactly the set of all points that are the same distance from $A$ and $B$ .

Worked example: Equal distances with algebra

Point $P$ lies on the perpendicular bisector of $\overline{AB}$ . If $PA = 3x + 2$ and $PB = 5x - 8$ , find $PA$ .

Solution. Points on the perpendicular bisector are equidistant from $A$ and $B$ , so $PA = PB$ :

\begin{aligned} 3x + 2 &= 5x - 8 \\ 10 &= 2x \\ x &= 5 \end{aligned}

$PA = 3(5) + 2 = 17$ . Check: $PB = 5(5) - 8 = 17$ .

Angle bisectors

An angle bisector is a ray that divides an angle into two congruent angles. The distance from a point to a line (or side) is always measured along the perpendicular segment, since that's the shortest path.

BP bisects angle ABC. The perpendicular distances PX and PY from P to the sides are equal.

Here $\triangle PXB \cong \triangle PYB$ by AAS: both have a right angle, $\angle PBX \cong \angle PBY$ because $\overrightarrow{BP}$ is a bisector, and they share the hypotenuse $\overline{BP}$ . So $PX = PY$ .

Angle Bisector Theorem and its converse

Theorem: If a point is on the bisector of an angle, then it is equidistant from the two sides of the angle.
Converse: If a point in the interior of an angle is equidistant from the two sides, then it is on the angle's bisector.

Worked example: Distance to the sides

Point $P$ is in the interior of $\angle ABC$ . The perpendicular distance from $P$ to $\overrightarrow{BA}$ is $2y + 1$ , and the perpendicular distance from $P$ to $\overrightarrow{BC}$ is $y + 6$ . If $\overrightarrow{BP}$ bisects $\angle ABC$ , how far is $P$ from each side?

Solution. By the Angle Bisector Theorem, the distances are equal:

2y + 1 = y + 6 \quad\Longrightarrow\quad y = 5.

The distance is $2(5) + 1 = 11$ units to each side.

Common mistake

"Distance to a side" always means the perpendicular distance. A slanted segment from $P$ to a side is longer than the perpendicular one, and the Angle Bisector Theorem says nothing about it. Look for the right-angle marks before you set two lengths equal.

Where the bisectors of a triangle meet

When three or more lines pass through one point, they are concurrent, and that point is the point of concurrency. A triangle's three perpendicular bisectors are always concurrent, and so are its three angle bisectors.

The circumcenter

Let $P$ be the point where the perpendicular bisectors of $\overline{AB}$ and $\overline{BC}$ cross. Then $PA = PB$ (from the first bisector) and $PB = PC$ (from the second). So $PA = PC$ , which puts $P$ on the third perpendicular bisector too. Because $P$ is the same distance from all three vertices, a circle centered at $P$ passes through $A$ , $B$ and $C$ .

Definition

Circumcenter

The circumcenter of a triangle is the point where its three perpendicular bisectors meet. It is equidistant from the three vertices, so it is the center of the circle that passes through all three vertices (the circumscribed circle).

The circumcenter is inside an acute triangle, at the midpoint of the hypotenuse of a right triangle, and outside an obtuse triangle.

Worked example: Finding a circumcenter with coordinates

Find the circumcenter of the triangle with vertices $A(0, 0)$ , $B(6, 0)$ and $C(2, 4)$ .

The perpendicular bisectors of the sides meet at P(3, 1), the center of the circle through A, B and C.Open in grapher →

Solution. You only need two perpendicular bisectors.

$\overline{AB}$ is horizontal with midpoint $(3, 0)$ , so its perpendicular bisector is the vertical line $x = 3$ .
$\overline{AC}$ has midpoint $(1, 2)$ and slope $\dfrac{4 - 0}{2 - 0} = 2$ . The perpendicular slope is $-\dfrac{1}{2}$ , so its bisector is $y - 2 = -\dfrac{1}{2}(x - 1)$ .

Substitute $x = 3$ : $y - 2 = -\dfrac{1}{2}(2) = -1$ , so $y = 1$ . The circumcenter is $(3, 1)$ .

Check with the Distance Formula: $PA = \sqrt{3^2 + 1^2} = \sqrt{10}$ , $PB = \sqrt{3^2 + 1^2} = \sqrt{10}$ and $PC = \sqrt{1^2 + 3^2} = \sqrt{10}$ . All equal.

The incenter

The same reasoning works for angle bisectors. The point where two angle bisectors meet is equidistant from all three sides, so it lies on the third bisector as well.

Definition

Incenter

The incenter of a triangle is the point where its three angle bisectors meet. It is equidistant from the three sides, so it is the center of the largest circle that fits inside the triangle (the inscribed circle). The incenter is always inside the triangle.

The angle bisectors (dashed) of right triangle ABC meet at the incenter I. The inscribed circle touches all three sides.

Worked example: Distances from the incenter

$I$ is the incenter of $\triangle ABC$ . The perpendicular from $I$ to $\overline{AB}$ meets it at $D$ , and $ID = 5$ . If $IA = 13$ , find $AD$ . How far is $I$ from side $\overline{BC}$ ?

Solution. $\triangle ADI$ has a right angle at $D$ , with hypotenuse $IA = 13$ . By the Pythagorean Theorem,

AD = \sqrt{13^2 - 5^2} = \sqrt{169 - 25} = \sqrt{144} = 12.

The incenter is equidistant from all three sides, so its distance to $\overline{BC}$ is also $5$ .

Tip

Keep the two centers straight with their names: the circumcenter is tied to the circle around the triangle (through the vertices), and the incenter is tied to the circle inside (touching the sides).

Practice

Practice 1

Point $P$ lies on the perpendicular bisector of $\overline{AB}$ , and $PA = 14$ . What is $PB$ ?

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 2

A city wants to build a fire station that is the same distance from three neighborhoods, located at the vertices of a triangle. Which point should it use?

Practice 3

Point $C$ lies on the perpendicular bisector of $\overline{AB}$ . If $CA = 4x - 3$ and $CB = 2x + 9$ , what is $CA$ ?

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 4

$\overrightarrow{BD}$ bisects $\angle ABC$ . If $m\angle ABD = (3x + 4)^\circ$ and $m\angle DBC = (5x - 10)^\circ$ , what is $m\angle ABC$ , in degrees?

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 5

A right triangle has vertices $(0, 0)$ , $(10, 0)$ and $(0, 24)$ . What are the coordinates of its circumcenter?

Enter a point like (2, -3)

Practice 6

$I$ is the incenter of $\triangle ABC$ . The perpendicular from $I$ to $\overline{AC}$ meets $\overline{AC}$ at $E$ . If $IE = 6$ and $IA = 10$ , what is $AE$ ?

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 7

Where is the circumcenter of an obtuse triangle?

Practice 8

Find the circumcenter of the triangle with vertices $A(0, 0)$ , $B(8, 0)$ and $C(2, 6)$ .

Enter a point like (2, -3)