Geometry

Perpendicular bisectors of a triangle

Which points are exactly as far from one point as from another, and is there a point that is equally far from three?

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Perpendicular bisector of a segmentMath 8

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CircumcenterGeometry

Where you are 4 in Special segments in a triangle

Geometry11

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Special segments in a triangle4

Medians and altitudes Angle bisectors of a triangle Midsegments of triangles and trapezoids

Perpendicular bisectors of a triangle

Why this exists

Two friends live at points A and B and want a meeting place that is fair: exactly as far from one home as from the other. The midpoint of \overline{AB} works. So does a point far off to the side, if it happens to be placed just right. There are infinitely many fair points, and testing them one at a time with a ruler would never end.

The perpendicular bisector was defined by how it is built: the line through the midpoint, at a right angle. That definition says nothing about distance to A or to B. This lesson shows that the two ideas pick out the very same line. Every point of the perpendicular bisector is equally far from A and B, and no other point is. A question about distances, with infinitely many candidates, becomes a line you can draw.

Then add a third friend. A triangle has three sides and so three perpendicular bisectors, and the same fact shows they all pass through one point: the one place equally far from all three corners.

Definition

A point P is equidistant from A and B when

PA = PB

Here PA is the length of \overline{PA}, the distance from P to A.

Figure 1Every point of \ell, such as P, is as far from A as from B.

The perpendicular bisectors of a triangle are the perpendicular bisectors of its three sides. Each one is a whole line through the midpoint of its side, at a right angle to that side. It belongs to the side, not to the opposite corner, and in general it does not pass through any vertex.

Intuition

Fold the paper so that A lands on B. The crease is the perpendicular bisector \ell of \overline{AB}. A point P on the crease does not move when the paper folds, while A moves onto B. So the segment \overline{PA} lands exactly on \overline{PB}, and the two are the same length. That is the whole theorem, seen in one fold.

Now take a point that is not on the crease, say on A’s side of it. To walk from it to B you have to cross the crease somewhere, and crossing it is a detour compared with going straight to A. So a point off the line is always nearer the endpoint on its own side.

The line \ell cuts the plane into two halves. In A’s half every point is nearer to A; in B’s half every point is nearer to B; and the line itself is the border where the two distances tie.

Results

Perpendicular Bisector Theorem. If a point lies on the perpendicular bisector of \overline{AB}, then it is equidistant from A and B.

Converse. If a point is equidistant from A and B, then it lies on the perpendicular bisector of \overline{AB}.

With \ell the perpendicular bisector of \overline{AB}, the two together say

PA = PB \iff P \text{ is on } \ell

so the perpendicular bisector is exactly the set of points equidistant from A and B, its locus.

Off the line. A point that is not on \ell is strictly nearer to the endpoint on its own side of \ell.

Two points settle it. If two different points are each equidistant from A and B, the line through them is the perpendicular bisector of \overline{AB}.

The perpendicular bisectors of a triangle meet. In every triangle ABC the three perpendicular bisectors pass through one point O, and

OA = OB = OC

O is the only point equidistant from all three vertices. It is called the circumcenter, because the circle centered at O through A also passes through B and C.

Figure 2The three perpendicular bisectors of a triangle pass through one point, O.

Why it is true

The theorem. Let \ell be the perpendicular bisector of \overline{AB}, crossing it at M, and let P be a point of \ell. If P = M, then PA = MA = MB = PB because M is the midpoint. Otherwise P is off the line AB, and \overline{PM} splits the figure into \triangle PMA and \triangle PMB (Figure 1).

Statements	Reasons
MA = MB	M is the midpoint of \overline{AB}
m\angle PMA = m\angle PMB = 90^\circ	\ell is perpendicular to \overline{AB}
PM = PM	the same segment (reflexive property)
\triangle PMA \cong \triangle PMB	SAS
PA = PB	corresponding parts of congruent triangles

The converse. Suppose PA = PB, and let M be the midpoint of \overline{AB}.

If P is on the line AB, it cannot lie outside the segment: beyond B, say, it would have PA = PB + BA, which is more than PB. So P is between A and B with PA = PB, which makes P the midpoint M, and M is on \ell.

If P is off the line AB, draw \overline{PM}:

Statements	Reasons
PA = PB	given
MA = MB	M is the midpoint of \overline{AB}
PM = PM	the same segment (reflexive property)
\triangle PMA \cong \triangle PMB	SSS
m\angle PMA = m\angle PMB	corresponding parts of congruent triangles
m\angle PMA + m\angle PMB = 180^\circ	they form a linear pair along \overline{AB}
m\angle PMA = 90^\circ	two equal angles adding to 180^\circ

So the line PM passes through the midpoint of \overline{AB} at a right angle. A segment has exactly one such line, so line PM is \ell, and P lies on it.

Off the line. Let P be on A’s side of \ell but not on \ell, as in Figure 3.

Figure 3The path from P to B has to cross \ell at some point Q, and QB = QA.

P and B are on opposite sides of \ell, so \overline{PB} crosses \ell at some point Q. By the theorem, QB = QA, so

PB = PQ + QB = PQ + QA

By the triangle inequality, PQ + QA \ge PA, with equality only if Q lies on \overline{PA}. It does not: both ends of \overline{PA} are on A’s side of \ell, so the whole segment is, and it never touches \ell. Hence PB > PA.

Two points settle it. By the converse, both points lie on the perpendicular bisector. Through two different points there is only one line, so that line is the perpendicular bisector.

The triangle. Call the perpendicular bisectors of \overline{AB} and \overline{BC} \ell_1 and \ell_2. First, they are not parallel. If they were, line AB, which is perpendicular to \ell_1, would also be perpendicular to \ell_2, since a line perpendicular to one of two parallel lines is perpendicular to the other. Then lines AB and BC would both pass through B and both be perpendicular to \ell_2, and through a point there is only one perpendicular to a line. They would be one line, and A, B, C would not make a triangle.

So \ell_1 and \ell_2 cross, at a point O. Because O is on \ell_1, OA = OB. Because O is on \ell_2, OB = OC. So OA = OC, and by the converse O lies on the perpendicular bisector of \overline{AC} as well. All three lines pass through O.

Any point equidistant from all three vertices lies on both \ell_1 and \ell_2, and two lines that cross meet only once, so O is the only such point.

Worked examples

Example 1

Lengths along the bisector

In Figure 1, line \ell is the perpendicular bisector of \overline{AB} at M, and P is on \ell. Suppose PA = 3x + 5, PB = 5x - 9 and AB = 20. Find x, PA and PM.

Solution

P lies on the perpendicular bisector, so by the theorem it is equidistant from A and B:

\begin{aligned} 3x + 5 &= 5x - 9 \\ 14 &= 2x \\ x &= 7 \end{aligned}

Then PA = 3(7) + 5 = 26 and PB = 5(7) - 9 = 26, which agree as they must.

For PM, use the right angle. M is the midpoint, so AM = \tfrac{1}{2}(20) = 10, and \angle PMA is a right angle because \ell \perp \overline{AB}. So \triangle PMA is a right triangle with hypotenuse \overline{PA}, and the Pythagorean theorem gives

PM = \sqrt{26^2 - 10^2} = \sqrt{676 - 100} = \sqrt{576} = 24

Example 2

Testing points on a grid

Let A = (1, 2) and B = (7, 4). Decide whether P = (3, 6) and Q = (5, 1) lie on the perpendicular bisector of \overline{AB}.

Solution

The theorem and its converse together say a point is on the perpendicular bisector exactly when it is equidistant from A and B. So measure, with the distance formula. Distances are never negative, so two distances are equal exactly when their squares are, and the squares avoid square roots.

\begin{aligned} PA^2 &= (3 - 1)^2 + (6 - 2)^2 = 4 + 16 = 20 \\ PB^2 &= (3 - 7)^2 + (6 - 4)^2 = 16 + 4 = 20 \end{aligned}

PA = PB, so P lies on the perpendicular bisector.

\begin{aligned} QA^2 &= (5 - 1)^2 + (1 - 2)^2 = 16 + 1 = 17 \\ QB^2 &= (5 - 7)^2 + (1 - 4)^2 = 4 + 9 = 13 \end{aligned}

QA \ne QB, so Q does not. It is nearer to B, so it lies on B’s side of the line.

Check. The midpoint of \overline{AB} is M = (4, 3). From A to B is 6 right and 2 up, the same direction as 3 right and 1 up; from M to P is 1 left and 3 up, which is that move turned a quarter turn. So \overline{MP} is square to \overline{AB}, as it should be.

Example 3

Two equidistant points fix the line

In Figure 4, CA = CB = 10, DA = DB = 17 and AB = 16, and the diagonals \overline{AB} and \overline{CD} cross at E. Show that line CD is the perpendicular bisector of \overline{AB}, then find AE, CE, DE and CD.

Figure 4C and D are each as far from A as from B.

Solution

C is equidistant from A and B, so by the converse it lies on the perpendicular bisector of \overline{AB}. So does D, for the same reason. C and D are different points, and through two points there is only one line, so line CD is the perpendicular bisector.

It crosses \overline{AB} at E, so E is the midpoint and the angles at E are right angles:

AE = \tfrac{1}{2}(16) = 8

In the right triangle AEC, the hypotenuse is CA = 10:

CE = \sqrt{10^2 - 8^2} = 6

In the right triangle AED, the hypotenuse is DA = 17:

DE = \sqrt{17^2 - 8^2} = 15

The diagonals cross, so E lies between C and D, and CD = 6 + 15 = 21.

Nothing in the given lengths mentioned a midpoint or a right angle. Two equal-distance facts supplied both.

Example 4

Three bisectors, one point

In \triangle ABC, the perpendicular bisectors of \overline{AB} and \overline{BC} meet at O, with OA = 4x + 1 and OC = 6x - 11.

(a) Find x and OB.

(b) Explain why the perpendicular bisector of \overline{AC} also passes through O.

Solution

(a) O is on the perpendicular bisector of \overline{AB}, so OA = OB. It is on the perpendicular bisector of \overline{BC}, so OB = OC. Together, OA = OC:

\begin{aligned} 4x + 1 &= 6x - 11 \\ 12 &= 2x \\ x &= 6 \end{aligned}

So OA = 4(6) + 1 = 25, and OB = OA = 25.

(b) OA = OC, so O is equidistant from A and C, and by the converse it lies on the perpendicular bisector of \overline{AC}. No third measurement was needed.

(c) N is the midpoint of \overline{AC}, so AN = 7, and \angle ONA is a right angle. In the right triangle ONA the hypotenuse is OA = 25:

ON = \sqrt{25^2 - 7^2} = 24

Example 5

The point equally far from three points

Find the point equidistant from A = (0, 0), B = (8, 0) and C = (2, 6), and the common distance.

Solution

The point lies on all three perpendicular bisectors, and since the three meet at one point, any two of them are enough.

\overline{AB} runs level from (0, 0) to (8, 0), so its midpoint is (4, 0) and its perpendicular bisector is the upright line x = 4. The point is therefore (4, y) for some y.

It must also be equidistant from A and C. Set the squared distances equal:

\begin{aligned} 4^2 + y^2 &= (4 - 2)^2 + (y - 6)^2 \\ 16 + y^2 &= 4 + y^2 - 12y + 36 \\ 12y &= 24 \\ y &= 2 \end{aligned}

The point is O = (4, 2).

Check. OA^2 = 4^2 + 2^2 = 20, OB^2 = (4 - 8)^2 + 2^2 = 20 and OC^2 = (4 - 2)^2 + (2 - 6)^2 = 20. All three agree, and the common distance is \displaystyle \sqrt{20} = 2\sqrt{5}. This is the triangle drawn in Figure 2.

Example 6

Isosceles triangles

Triangle ABC has AB = AC. Show that the perpendicular bisector of \overline{BC} passes through A, and that the segment from A to the midpoint of \overline{BC} is perpendicular to \overline{BC}.

Solution

AB = AC says A is equidistant from B and C. By the converse, A lies on the perpendicular bisector of \overline{BC}.

That bisector also passes through the midpoint N of \overline{BC}, by definition. A is not on \overline{BC}, since ABC is a triangle, so A \ne N, and the only line through both is line AN. So line AN is the perpendicular bisector, and the segment \overline{AN}, which lies along it, is perpendicular to \overline{BC}.

The theorem runs the other way too: if the perpendicular bisector of \overline{BC} passes through A, then AB = AC. So a triangle has AB = AC exactly when the perpendicular bisector of \overline{BC} goes through A. The segment \overline{AN} is the triangle’s median from A, and in an isosceles triangle it does double duty.

Example 7

A perimeter from two lengths

In \triangle ABC, AB = 8 and AC = 15. The perpendicular bisector of \overline{BC} meets side \overline{AC} at D, as in Figure 5. Find the perimeter of \triangle ABD.

Figure 5D lies on the perpendicular bisector of \overline{BC}, so DB = DC.

Solution

The perimeter is AB + BD + DA, and neither BD nor DA is given. But D lies on the perpendicular bisector of \overline{BC}, so by the theorem DB = DC. Replace one with the other:

AB + BD + DA = AB + DC + DA

D lies on \overline{AC} between A and C, so the two pieces add to the whole side: DC + DA = AC. The perimeter is

AB + AC = 8 + 15 = 23

The length BC never entered. The bisector carried \overline{BD} across onto \overline{DC}, which already sat on a side whose length was known.

Boundaries

Equidistant from two points, not from two lines. The distance from a point to a line is a different measurement, the shortest distance to any point of it. The points equidistant from the two sides of an angle lie on a different line, the angle bisector: that is the angle bisector distance theorem.

In a plane. In space, the points equidistant from A and B fill a whole plane, square to \overline{AB} through its midpoint. Every result here is about a flat figure.

One equidistant point does not locate the line. A single point with PA = PB is on the perpendicular bisector, but on its own it does not say where the line runs. A second point, which may be the midpoint, settles it.

Three points must not be on one line. Two points always have infinitely many equidistant points. Three points that form a triangle have exactly one. Three different points on one line have none, because the perpendicular bisectors of two of the gaps are both square to that line, pass through different midpoints, and so are different parallel lines that never meet.

The meeting point need not be inside the triangle. It is inside an acute triangle, on the longest side of a right triangle, and outside an obtuse one. The circumcenter takes those cases one at a time.

A side’s bisector usually misses the opposite corner. It passes through the opposite vertex only when the two sides meeting there are equal, as Example 6 showed.

Where this goes next

The meeting point of the three bisectors is the circumcenter, and the circle centered there through one vertex passes through all three. That circle, and how to draw it with compass and straightedge, rest entirely on this lesson.

The compass construction of a perpendicular bisector works because of the converse. Two arcs of the same radius, one centered at A and one at B, cross at two points, and each crossing is the same distance from A as from B. Two equidistant points settle the line, so the line through the crossings is the perpendicular bisector: see bisecting a segment with compass and straightedge.

The center of a circle is the same distance from every point on it, so it lies on the perpendicular bisector of every chord. That is the heart of the chord theorems, and it finds the center of any circle from two chords.

The twin of this lesson swaps points for lines: points equidistant from the two sides of an angle lie on its bisector, and the three angle bisectors of a triangle also meet at one point, in angle bisectors of a triangle. In a reflection, the mirror is the perpendicular bisector of the segment from each point to its image, which is why every point of the mirror is as far from a point as from its image. And the two-point argument of Example 3 is what shows that each diagonal of a rhombus lies on the perpendicular bisector of the other.

Exercises

Points on the bisector

In Exercises 1 to 5, line \ell is the perpendicular bisector of \overline{AB} and crosses it at M, and P is a point of \ell other than M.

PA = 11 and AB = 12. Find PB and PM.
PA = 5x - 4 and PB = 3x + 10. Find x and PA.
AB = 18 and PM = 12. Find PA.
PA = 41 and AB = 18. Find PM.
PA = x^2 and PB = 2x + 15. Find every value of x that works, and the length PA each one gives.

Points equally far from both ends

P and Q are different points with PA = PB = 7 and QA = QB = 4. What is line PQ? Explain.
A point P has PA = 12 and PB = 13. Does P lie on the perpendicular bisector of \overline{AB}? If not, which side of it is P on?
Quadrilateral ABCD has AB = AD = 13 and CB = CD = 20, and its diagonals cross at E. If BD = 24, find BE, AE, CE and AC.

On a grid

Let A = (2, 1) and B = (6, 5). Decide whether P = (1, 6) and Q = (5, 3) lie on the perpendicular bisector of \overline{AB}.
Let A = (1, 1) and B = (7, 3). Find the point on the x-axis that is equidistant from A and B.
Find the point equidistant from A = (0, 0), B = (6, 0) and C = (0, 8), and the common distance.
Find the point equidistant from A = (-1, 1), B = (5, 1) and C = (3, 5), and the common distance.

In a triangle

The perpendicular bisectors of \triangle ABC meet at O, with OA = 2x + 9 and OB = 5x - 3. Find x and OC.
The perpendicular bisectors of \triangle ABC meet at O, and OB = 10. The perpendicular bisector of \overline{BC} crosses \overline{BC} at N, and BC = 12. Find ON.
In \triangle ABC, AB = 11 and AC = 17. The perpendicular bisector of \overline{BC} meets side \overline{AC} at D, and DC = 10. Find DB, DA and the perimeter of \triangle ABD.
In \triangle PQR, the perpendicular bisector of \overline{QR} passes through P. If PQ = 3y - 2 and PR = y + 12, find y and PQ.

Reasoning

Explain why the midpoint is the only point of the line AB that is equidistant from A and B.
A rhombus has four equal sides. Show that each diagonal of a rhombus lies on the perpendicular bisector of the other.
\ell is the perpendicular bisector of \overline{AB}, and P is a point of \ell not on \overline{AB}. Explain why \triangle PAB is isosceles and why its angles at A and B are equal.
A, B and C are three different points on one line. Explain why no point is equidistant from all three.
Kai says the point where the three perpendicular bisectors of a triangle meet always lies inside the triangle. Test the claim with the triangle of Exercise 11.
Describe the set of all points that are nearer to A than to B, and explain why it is that set.

Answers

PB = 11 and \displaystyle PM = \sqrt{85}. P is on \ell, so PB = PA. AM = 6 and \angle PMA is right, so \displaystyle PM = \sqrt{11^2 - 6^2} = \sqrt{121 - 36} = \sqrt{85}.
x = 7 and PA = 31. By the theorem 5x - 4 = 3x + 10, so 2x = 14. Check: 5(7) - 4 = 31 = 3(7) + 10.
PA = 15. AM = 9, and in the right triangle PMA, \displaystyle PA = \sqrt{9^2 + 12^2} = \sqrt{225}.
PM = 40. AM = 9, so \displaystyle PM = \sqrt{41^2 - 9^2} = \sqrt{1681 - 81} = \sqrt{1600}.
x = 5, giving PA = 25, or x = -3, giving PA = 9. From x^2 = 2x + 15, x^2 - 2x - 15 = 0 and (x - 5)(x + 3) = 0. Both give a positive length: PB = 2(5) + 15 = 25 and PB = 2(-3) + 15 = 9.
Line PQ is the perpendicular bisector of \overline{AB}. Each of P and Q is equidistant from A and B, so each lies on that bisector by the converse, and two different points lie on only one line.
No, since 12 \ne 13, and P is on A’s side. A point on B’s side would be nearer to B, and a point on the line would be equally far from both, so the only place left is A’s side.
BE = 12, AE = 5, CE = 16 and AC = 21. A and C are each equidistant from B and D, so line AC is the perpendicular bisector of \overline{BD}, E is its midpoint and the angles at E are right. Then \displaystyle AE = \sqrt{13^2 - 12^2} = 5, \displaystyle CE = \sqrt{20^2 - 12^2} = 16, and AC = 5 + 16 = 21.
P is on it and Q is not. PA^2 = 1 + 25 = 26 and PB^2 = 25 + 1 = 26. QA^2 = 9 + 4 = 13 and QB^2 = 1 + 4 = 5, so Q is nearer to B.
\left(\tfrac{14}{3}, 0\right). For (x, 0): (x - 1)^2 + 1 = (x - 7)^2 + 9, so x^2 - 2x + 2 = x^2 - 14x + 58 and 12x = 56.
(3, 4), at distance 5. The bisector of \overline{AB} is x = 3 and the bisector of \overline{AC} is y = 4. Each distance is \displaystyle \sqrt{3^2 + 4^2} = 5.
(2, 2), at distance \displaystyle \sqrt{10}. The bisector of \overline{AB} is x = 2. For (2, y): 9 + (y - 1)^2 = 1 + (y - 5)^2, so 10 - 2y = 26 - 10y and 8y = 16. Then OA^2 = OB^2 = OC^2 = 10.
x = 4 and OC = 17. OA = OB, so 2x + 9 = 5x - 3 and 3x = 12. OC = OB = 5(4) - 3.
ON = 8. N is the midpoint of \overline{BC}, so BN = 6, and \angle ONB is right, so \displaystyle ON = \sqrt{10^2 - 6^2} = 8.
DB = 10, DA = 7 and the perimeter is 28. D is on the bisector of \overline{BC}, so DB = DC = 10; DA = AC - DC = 17 - 10 = 7; and 11 + 10 + 7 = 28, which is AB + AC.
y = 7 and PQ = 19. P is on the perpendicular bisector of \overline{QR}, so PQ = PR: 3y - 2 = y + 12 gives 2y = 14, and PQ = 3(7) - 2 = 19.
A point of line AB outside the segment, beyond B say, has PA = PB + AB > PB, and beyond A the same holds the other way round. A point between A and B with PA = PB splits the segment into two equal parts, which is what the midpoint is, and a segment has only one.
Let the rhombus be ABCD, with diagonals \overline{AC} and \overline{BD}. AB = AD and CB = CD, so A and C are each equidistant from B and D. By the converse both lie on the perpendicular bisector of \overline{BD}, and since A \ne C, that bisector is line AC. The same argument with B and D, each equidistant from A and C, shows line BD is the perpendicular bisector of \overline{AC}.
By the theorem PA = PB, so two sides are equal and the triangle is isosceles. The proof of the theorem showed \triangle PMA \cong \triangle PMB, and the angles at A and B are corresponding parts, so they are equal. (This is also the base angles theorem.)
A point equidistant from all three would lie on the perpendicular bisectors of \overline{AB} and \overline{BC}. Both are perpendicular to the one line through A, B and C, so they are parallel. They pass through different midpoints, so they are different lines, and different parallel lines never meet.
The claim is false. The bisectors meet at (3, 4), which is the midpoint of \overline{BC}, since \tfrac{1}{2}(6 + 0) = 3 and \tfrac{1}{2}(0 + 8) = 4. The point is on a side of the triangle, not inside it. (That triangle has a right angle at A.)
The half of the plane on A’s side of the perpendicular bisector of \overline{AB}, not including the line itself. Points on A’s side are strictly nearer to A by the off-the-line result; points on the line are equally far from both; points on B’s side are nearer to B.