Perpendicular Bisectors and the Circumcenter

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Learning goals

Prove both directions of the perpendicular-bisector theorem, including its midpoint case
Construct segment bisectors and perpendiculars with a compass and straightedge
Justify concurrence of perpendicular bisectors and construct a triangle circumcircle
Explain circumcenter locations for acute, right, and obtuse triangles
Recover a given rotation center from suitable point-image pairs

Equal distances describe a line

The perpendicular bisector of a segment is the line perpendicular to it through its midpoint. A segment bisector passes through the midpoint; it need not be perpendicular.

Take a segment 8 cm long. Its midpoint is 4 cm from each end. A point 3 cm away from the midpoint along the perpendicular makes two right triangles, each with legs 3 cm and 4 cm. Pythagoras gives a 5 cm hypotenuse on each side. Moving along that perpendicular changes both distances together.

The two triangles share one leg and have equal halves of the base.

For two distinct points $A$ and $B$ , the perpendicular-bisector theorem says that every point on their perpendicular bisector is equally far from them. Its converse says that every point equally far from $A$ and $B$ lies on that line.

The perpendicular-bisector theorem and its converse#

Let $M$ be the midpoint of $AB$ .

For the forward direction, first take $P$ on the perpendicular bisector with $P$ different from $M$ . Triangles $PMA$ and $PMB$ share $PM$ , have congruent halves $MA$ and $MB$ , and have a right included angle at $M$ . SAS makes them congruent, so their corresponding sides $PA$ and $PB$ are equal. If $P$ is $M$ itself, the equal distances follow directly from the midpoint definition; there are no two triangles to compare.

For the converse, suppose first that $P$ is off line $AB$ and its distances to the endpoints are equal. Triangles $PMA$ and $PMB$ have equal sides $PA$ and $PB$ , equal halves $MA$ and $MB$ , and common side $PM$ . SSS makes their angles at $M$ equal. Those angles form a linear pair, so each is a right angle. Thus $PM$ is the perpendicular bisector.

If $P$ is on line $AB$ , a point at or beyond either endpoint has unequal endpoint distances by segment addition. A point between the endpoints has equal distances only when it is their midpoint. The midpoint is on the perpendicular bisector.

The 3 cm and 4 cm lengths helped us see the result, but neither congruence argument depends on those lengths.

A locus is the whole set of points satisfying a condition. Both directions matter when describing a locus: the perpendicular bisector is exactly the locus of points equally far from two distinct endpoints, with no missing points and no extra ones. If the endpoints coincide, every point has equal distances to them, so the locus is the whole plane instead.

Worked example 1 Use equidistance before algebra

Problem. $M$ is the midpoint of $AB$ , and $PM$ is perpendicular to $AB$ . The lengths of $PA$ and $PB$ , in centimeters, are given below. Find $x$ and both lengths.

\begin{gathered} PA=3x+2\\ PB=5x-12 \end{gathered}

M is the midpoint of AB, and PM is perpendicular to AB.

Solution. $P$ lies on the perpendicular bisector, so the two expressions measure equal lengths.

\begin{aligned} 3x+2&=5x-12\\ 14&=2x\\ x&=7 \end{aligned}

Substitute into each expression separately.

\begin{aligned} PA&=3(7)+2\\ PA&=23\text{ cm}\\ PB&=5(7)-12\\ PB&=23\text{ cm} \end{aligned}

Check your understanding

A point Q is equally far from distinct endpoints A and B. Which conclusion follows, even if Q lies on AB?

A Q is an endpoint of AB B Q lies on the perpendicular bisector of AB C Q must be off line AB D Q is equally far from every point of AB

Build perpendiculars from equal radii

A compass can create equal distances without reading a ruler. To bisect a segment, draw two circles centered at its endpoints with the same radius, greater than half the segment length. For an 8 cm segment, a 5 cm opening works; a 4 cm opening produces only one touching point.

Keep the compass opening unchanged; join both crossings.

Call the two distinct crossings $X$ and $Y$ . Each crossing is equally far from $A$ and $B$ , since the circles have equal radii. The converse puts both crossings on the perpendicular bisector. Two distinct points determine one line, so drawing line $XY$ constructs it. Its crossing with $AB$ constructs the midpoint and therefore also bisects the segment.

Worked example 2 Draw a perpendicular at a point

Problem. A point $P$ lies on a given line. Construct a perpendicular to the line at $P$ using a compass and unmarked straightedge, and justify the construction.

Solution. Draw a circle centered at $P$ with any positive radius. Let its intersections with the line be $A$ and $B$ . They lie on opposite sides of $P$ with equal distances from it, so $P$ is their midpoint.

From $A$ and $B$ , draw equal-radius arcs with an opening greater than half $AB$ . Choose a crossing $X$ off the line. Draw $PX$ .

The first circle selects A and B; equal-radius arcs then determine the perpendicular.

The converse puts $X$ on the perpendicular bisector of $AB$ . That bisector also passes through the known midpoint $P$ , so line $PX$ is perpendicular at $P$ .

For a perpendicular from a point off a line, begin with a circle centered at that point $P$ . Choose its radius large enough to cut the line in two distinct points $A$ and $B$ . Then $P$ is equally far from them. Circles centered at $A$ and $B$ with radius $AP$ already pass through $P$ and meet again at a second point $X$ , on the other side of the line. The strict triangle inequality in nondegenerate triangle $APB$ ensures that this radius is greater than half $AB$ .

The first circle selects A and B; equal-radius arcs then determine the perpendicular.

Both $P$ and $X$ are on the perpendicular bisector of $AB$ , so $PX$ is the required perpendicular. Its crossing with the given line is the perpendicular foot. Extend a given segment to its supporting line if necessary. A circle that merely touches the line supplies only one point and cannot start this construction.

Check your understanding

For distinct endpoints A and B, which equal compass radius guarantees two distinct circle crossings?

A Exactly half AB B Any positive radius C Greater than half AB D Less than half AB

One circle through three vertices

Two equal-distance conditions can locate a single point. If a point is 7 cm from both $A$ and $B$ , and also 7 cm from $C$ , it can be the center of a circle through all three. The issue is finding that point without guessing the radius.

Lines are concurrent when they pass through one common point. In every nondegenerate triangle, its three perpendicular bisectors are concurrent. Their common point is the circumcenter, and the circle centered there through the three vertices is the circumcircle. The common vertex distance is the circumradius. These distances go to vertices, not to side lines.

Why the three perpendicular bisectors concur#

Construct the perpendicular bisectors of $AB$ and $AC$ . They cannot be parallel: two lines perpendicular to parallel lines are parallel, so parallel bisectors would force $AB$ and $AC$ to be parallel. Those sides meet at $A$ and are distinct because the triangle is nondegenerate. The two bisectors therefore meet at one point $O$ .

The forward theorem applied to the first bisector gives

OA=OB.

Applied to the second, it gives

OA=OC.

Thus $OB$ and $OC$ are equal. The converse places $O$ on the perpendicular bisector of $BC$ . All three pass through $O$ . Any other point equally far from all three vertices would have to lie on the first two bisectors, whose intersection is unique.

Use Back to return to the triangle, then Next to build the perpendicular bisectors. Drag vertex A or change the base angles while BC stays fixed at 6 units. Before choosing a right or obtuse case, predict which side of the triangle will contain O. The displayed measurements illustrate the proof above.

Three perpendicular bisectors locate one equal-distance point

Construction: complete construction

Scroll sideways to see the whole diagram.

Angle B Angle C Construction Try a case

Angle B = 55°. Angle C = 65°. Angle A = 60°. The triangle is acute. The three perpendicular bisectors meet at O. OA ≈ 3.46 units. OB ≈ 3.46 units. OC ≈ 3.46 units. O is inside the triangle. The circumcircle passes through A, B and C.

The same construction works for acute, right and obtuse triangles. Compare the three vertex distances and the circle through A, B and C.

To construct the circumcircle, keep the first two bisectors, set the compass to $OA$ , and draw the circle centered at $O$ . The proof establishes that it also passes through $B$ and $C$ . The third bisector is a useful check, but is not needed to locate $O$ .

Worked example 3 Connect a radius to a right triangle

Problem. $O$ is the circumcenter of triangle $ABC$ . $M$ is the midpoint of $AB$ , whose length is 24 cm. The length of $OM$ is 5 cm. Find the circumradius and explain why the same radius reaches $C$ .

O is the circumcenter, AB is 24 cm long, and OM is 5 cm long.

Solution. The bisector through $M$ is perpendicular to $AB$ . Half of the 24 cm side is 12 cm, so triangle $OMA$ is right with legs 5 cm and 12 cm.

\begin{aligned} OA^2&=5^2+12^2\\ OA^2&=169\\ OA&=13\text{ cm} \end{aligned}

By the circumcenter theorem, $OC$ has the same length as $OA$ . Thus the circle with center $O$ and radius 13 cm reaches every vertex.

Check your understanding

O lies on the perpendicular bisectors of AB and AC in a nondegenerate triangle. Which statement completes the proof for the third side?

A AB and AC have equal lengths B O lies inside the triangle C The three sides are parallel D OB and OC are equal, so the converse puts O on the bisector of BC

Where the center lies

A triangle’s circumcenter need not lie inside it. In a triangle with angles 50, 60, and 70 degrees it is inside. If the largest angle opens to 90 degrees, the center reaches the opposite side’s midpoint. If it opens past 90 degrees, the center crosses that side to the outside.

Here is a justification using midpoint parallels and triangle angles. Name a longest side $AB$ , so the other vertex is $C$ . The angles at $A$ and $B$ are acute, since the largest angle is at $C$ . Let $M$ and $N$ be the midpoints of $AB$ and $AC$ . The midsegment theorem makes $MN$ parallel to $BC$ . Thus triangle $AMN$ has the same three angles as triangle $ABC$ , in the same order.

Circumcenter locations from midpoint parallels#

Write the angles at $A$ , $B$ , and $C$ as $\alpha$ , $\beta$ , and $\gamma$ . In triangle $AMN$ , the angles at $A$ , $M$ , and $N$ have those respective measures. The circumcenter is the intersection of the perpendicular at $M$ to $AM$ and the perpendicular at $N$ to $AN$ .

The perpendicular foot from $N$ to $AB$ is between $A$ and $M$ , because both endpoint angles of $ABC$ are acute and $N$ is halfway along $AC$ . Draw the ray through $N$ parallel to $AB$ toward $B$ . It reaches the perpendicular through $M$ . The ray from $N$ perpendicular to $AN$ on that same side of $AN$ makes an acute angle of $90^\circ-\alpha$ below the parallel ray. The same-side interior angles with the perpendicular through $M$ sum to less than 180 degrees, so the two rays meet on that side by the parallel postulate. Call their intersection $O$ . At $N$ , ray $NO$ makes 90 degrees with $NA$ .

If $\gamma$ is 70 degrees, the perpendicular ray lies beyond $NM$ by 20 degrees, across line $MN$ from $A$ . It therefore meets the perpendicular through $M$ on the same side of $AB$ as $N$ and $C$ . In general, this gap is $90^\circ-\gamma$ when $\gamma$ is acute. Equivalently, an intersection below $AB$ would make triangle $MNO$ have angle $90^\circ+\beta$ at $M$ and angle $\gamma-90^\circ$ at $N$ , which is impossible for acute $\gamma$ .

If $\gamma$ is 90 degrees, $NM$ itself is perpendicular to $AN$ , so the intersection is $M$ .

If $\gamma$ is 110 degrees, the perpendicular ray lies inside angle $ANM$ , 20 degrees short of $NM$ . It meets the perpendicular through $M$ across $AB$ from $C$ . In general, that gap is $\gamma-90^\circ$ when $\gamma$ is obtuse. An intersection on $C$ ‘s side would instead require the negative angle $90^\circ-\gamma$ at $N$ .

For an acute triangle, the same argument can start with each of its three sides, since all endpoint angles are acute. $O$ is therefore on the vertex side of every supporting side line. The intersection of those three open half-planes is exactly the triangle’s interior. For an obtuse triangle, being across the longest side from $C$ already puts $O$ outside the triangle. In the right case, $M$ is on its hypotenuse.

Acute case: O is inside all three side half-planes.

Right case: O is the midpoint of hypotenuse AB.

Obtuse case: O is across AB from the obtuse vertex C.

Worked example 4 Recognize the right-triangle center

Problem. A right triangle has legs 9 cm and 12 cm. Locate its circumcenter and find its circumradius.

The right angle is at A; the given legs are 9 cm and 12 cm.

Solution. Its hypotenuse length follows from Pythagoras.

\begin{aligned} c^2&=9^2+12^2\\ c^2&=225\\ c&=15\text{ cm} \end{aligned}

The right-case proof puts the circumcenter at the hypotenuse midpoint. Its distance to a hypotenuse endpoint is half 15 cm, so the circumradius is 7.5 cm. This conclusion depends on the triangle being right.

Check your understanding

A triangle has an obtuse angle at C. Where is its circumcenter?

A Across AB from C, outside the triangle B At C C At the midpoint of AC D Always inside the triangle

Recover a rotation center

A rotation keeps each point’s distance from its center. For example, if a 90 degree rotation carries a point 6 cm from the center to its image, the image remains 6 cm from the center. The center is therefore equally far from the two endpoints of their point-image chord [the segment joining the point to its image]. The converse puts the center on that chord’s perpendicular bisector.

Worked example 5 Justify the hidden-center construction

Problem. One given nonidentity rotation carries $A$ to $A'$ and $B$ to $B'$ . Both point-image segments have nonzero length and are not parallel. Construct the center and justify why the construction recovers that rotation’s center.

Solution. Construct the perpendicular bisector of $AA'$ , then that of $BB'$ . Call their intersection $O$ .

Each bisector contains the already-given rotation center.

The given rotation has a center equally far from $A$ and $A'$ , so that center belongs to the first bisector. It is also equally far from $B$ and $B'$ , so it belongs to the second. Nonparallel chords have nonparallel perpendicular bisectors, which meet at exactly one point. The given center must therefore be $O$ .

A fixed point produces a zero-length chord and cannot supply a bisector. Two unsuitable chords can have the same bisector, so choose another point-image pair if that occurs. For pairs known to come from one rotation, distinct chord bisectors already share its center; hence they cannot be distinct parallel lines.

This argument uses the fact that one rotation was given. Equal radii for two arbitrary pairs do not prove that their directed turn angles agree. Check the angle and direction as well if the task is to decide whether arbitrary pairs describe one rotation.

Check your understanding

Two arbitrary point pairs have equal endpoint distances from O. What extra check is needed before claiming one rotation about O maps both pairs?

A Their endpoint distances must all equal each other B Both chords must be diameters C The directed turn angles must agree D O must lie inside a triangle

Perpendicular Bisectors and the Circumcenter

Learning goals

Equal distances describe a line

Build perpendiculars from equal radii

One circle through three vertices

Where the center lies

Recover a rotation center

Common mistakes

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Multiple Choice Questions (MCQ)

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