Core practice ← Back to lesson

Perpendicular Bisectors and the Circumcenter: Core practice

10 practice problems for this lesson. Work on paper, use hints when you need them, and check the answer or the full solution when you are ready.

Difficulty: Core (core-course level)

0 of 10 done

Progress saved in this browser.

Problem 1 of 10

Problem 1 Connect equal lengths to a right triangle

M is the midpoint of AB, and PM is perpendicular to AB. The length of AB is 30 cm, and PM is 8 cm.

The length of PB, in centimeters, is $2x+5$ . Find PA, PB, and x. Explain the equal-distance step.

The stated side and perpendicular distance are drawn to one scale.

Text description of this figure

M is the midpoint of AB. AB is 30 centimeters long. PM is 8 centimeters long and is perpendicular to AB. P is off AB. No requested value is drawn.
1. Hint 1
  Identify the geometric condition that connects PA and PB.
2. Hint 2
  Use half of AB with PM in right triangle PMA, then compare PB with its given expression.
Answer

$\begin{gathered}PA=17\text{ cm}\\PB=17\text{ cm}\\x=6.\end{gathered}$

Full solution

P lies on the perpendicular bisector, so PA and PB have equal lengths.
The midpoint gives
$AM=15\text{ cm}.$

Pythagoras in right triangle PMA gives
$\begin{aligned}PA^2&=15^2+8^2\\PA^2&=289\\PA&=17\text{ cm}.\end{aligned}$

Therefore PB is also 17 cm.
Its expression gives
$\begin{aligned}2x+5&=17\\2x&=12\\x&=6.\end{aligned}$

Answer

$\begin{gathered}PA=17\text{ cm}\\PB=17\text{ cm}\\x=6.\end{gathered}$

Key idea

Combine an equidistance theorem with a right-triangle calculation.
Problem 2 Use a common radius and a half-side

O is the circumcenter of triangle ABC. Its distances to A and B, in centimeters, are given by $2y+3$ and $4y-11$ , respectively.

Side AB is 16 cm long. M is its midpoint. Find y, the circumradius, and OM.

O is the circumcenter, and M is the midpoint of the given side AB.

Text description of this figure

O is the circumcenter of triangle ABC. M is the midpoint of side AB, which is 16 centimeters long. No unknown numerical lengths are labeled.
1. Hint 1
  The circumcenter has equal distances to the vertices.
2. Hint 2
  Find the common radius first. Triangle OMA has that radius as hypotenuse and half of AB as one leg.
Answer

$\begin{gathered}y=7\\R=17\text{ cm}\\OM=15\text{ cm}.\end{gathered}$

Full solution

Equal vertex distances give
$\begin{aligned}2y+3&=4y-11\\14&=2y\\y&=7.\end{aligned}$

Substituting gives
$\begin{aligned}R&=2(7)+3\\R&=17\text{ cm}.\end{aligned}$

The side midpoint gives
$AM=8\text{ cm}.$
OM is perpendicular to AB because O is on its perpendicular bisector.

Pythagoras gives
$\begin{aligned}OM^2&=17^2-8^2\\OM^2&=225\\OM&=15\text{ cm}.\end{aligned}$

The solution establishes that OM is perpendicular to AB.

Answer

$\begin{gathered}y=7\\R=17\text{ cm}\\OM=15\text{ cm}.\end{gathered}$

Key idea

A circumradius reaches a vertex; the midpoint distance is a different length.
Problem 3 Prove the forward direction, including the midpoint

A and B are distinct. M is the midpoint of AB. A point P lies on the perpendicular to AB at M. Prove that PA and PB have equal lengths.

Your proof must cover P off AB and P equal to M separately.

M is the midpoint of AB, and P lies on the perpendicular at M.

Text description of this figure

A and B are distinct endpoints. M is their midpoint. P lies on the perpendicular through M. The connector triangles share PM.
1. Hint 1
  For a point off AB, look for two triangles with the midpoint halves and the shared connector.
2. Hint 2
  Use SAS when the triangles are nondegenerate. At the midpoint, use its definition directly.
Answer

PA and PB are equal for every P on the perpendicular bisector, including the midpoint.

Full solution

If P differs from M, triangles PMA and PMB share PM, have equal sides MA and MB, and have equal included right angles.
They are congruent by SAS.

Corresponding sides PA and PB are therefore equal.

If P is M, the distances are MA and MB.
They are equal by the midpoint definition; no triangle-congruence claim is needed.

Answer

PA and PB are equal for every P on the perpendicular bisector, including the midpoint.

Key idea

Handle a point that collapses a triangle directly instead of applying a triangle test.
Problem 4 Construct and annotate a segment bisector

Copy a nonzero segment AB on paper. Construct its perpendicular bisector with a compass and unmarked straightedge.

Label both compass crossings X and Y and the resulting midpoint M. Add the equality and right-angle marks your reasoning justifies. Explain the compass-opening condition and why the resulting line is correct.

Segment AB has distinct endpoints.

Text description of this figure

Segment AB has distinct endpoints. The midpoint, circle crossings, and perpendicular bisector are not supplied.
1. Hint 1
  Create points equally far from both endpoints.
2. Hint 2
  Choose one radius greater than half AB and keep it for the circles centered at A and B.
3. Hint 3
  Explain why both crossings belong to the desired line before marking the midpoint and right angle.
Answer

Draw equal-radius circles with radius greater than half AB; join their two crossings X and Y. Their line is the perpendicular bisector, and its crossing with AB is M.

Full solution

Draw circles centered at A and B with the same radius greater than half AB.
They meet in two distinct points X and Y on opposite sides of AB.

Each crossing has equal endpoint distances because both are radii of the equal circles.
By the converse, X and Y lie on the perpendicular bisector of AB.

Two distinct points determine one line, so XY is that bisector.
Its crossing M with AB is the midpoint.

Mark equal lengths AM and MB and a right angle where XY meets AB.
A radius equal to half AB would give only one touching point, and a smaller radius would give no crossing.

The completed construction has equal halves and a right angle at M.

Answer

Draw equal-radius circles with radius greater than half AB; join their two crossings X and Y. Their line is the perpendicular bisector, and its crossing with AB is M.

Key idea

Compass marks are justified by equal radii and the converse theorem.
Problem 5 Prove the converse without assuming the conclusion

P is off line AB, and its distances PA and PB are equal. M is the midpoint of AB. Prove that PM is perpendicular to AB and state what this says about the locus of such points P.

PA and PB are equal, and M is the midpoint of AB.

Text description of this figure

P is off line AB and has equal distances PA and PB. M is the midpoint of AB. No perpendicularity may be assumed in the proof.
1. Hint 1
  Use all three pairs of corresponding sides in triangles PMA and PMB.
2. Hint 2
  After congruence gives equal angles at M, use the fact that they form a linear pair.
Answer

PM is perpendicular to AB, so every such P lies on the perpendicular bisector of AB.

Full solution

Triangles PMA and PMB have equal sides PA and PB by hypothesis, equal halves MA and MB by the midpoint definition, and common side PM.
They are congruent by SSS.

Their angles at M are therefore equal.
They form a linear pair because MA and MB are opposite rays.

If each angle has measure t degrees, then
$\begin{aligned}2t&=180\\t&=90.\end{aligned}$
Thus PM is perpendicular to AB.

Every point off AB satisfying the equal-distance condition lies on the perpendicular bisector.
The midpoint is the only satisfying point on AB, and it lies on that same line, so no equal-distance points are missing from the locus.

Answer

PM is perpendicular to AB, so every such P lies on the perpendicular bisector of AB.

Key idea

The converse needs a side-based congruence test before any right angle is known.
Problem 6 Construct a perpendicular at the specified point

P is a given point on line l. Construct a perpendicular to l at P with compass and unmarked straightedge.

Label the two points used to make P a midpoint, and justify every equality or right angle you mark.

The point on the line is given.

Text description of this figure

The point P lies on line l. No other points, compass arcs, or perpendicular lines have been constructed.
1. Hint 1
  You need a segment on l whose midpoint is already known to be P.
2. Hint 2
  A circle centered at P makes two equally distant points on opposite sides of P.
3. Hint 3
  Use equal-radius arcs from those points to produce a second point on their perpendicular bisector.
Answer

Use a circle centered at P to mark A and B on l, then equal-radius arcs centered at A and B with radius greater than PA. Join P to an off-line crossing X.

Full solution

Draw any positive-radius circle centered at P.
Its line intersections A and B are on opposite sides of P, and PA and PB are equal radii, so P is the midpoint.

From A and B draw equal-radius arcs with a radius greater than PA, which is half AB.
They have distinct off-line crossings; choose one and name it X.

X is equally far from A and B, so the converse puts X on their perpendicular bisector.
P also lies on it because P is the midpoint.

The line PX is therefore that perpendicular bisector.
Mark PA equal to PB and a right angle at P.

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

Answer

Use a circle centered at P to mark A and B on l, then equal-radius arcs centered at A and B with radius greater than PA. Join P to an off-line crossing X.

Key idea

First turn the specified point into the midpoint of a convenient auxiliary segment.
Problem 7 Explain all three location cases

Case A: A triangle has angles 35, 65, and 80 degrees. Locate its circumcenter relative to the triangle and justify your answer.

Case B: A triangle has angles 40, 50, and 90 degrees. Locate its circumcenter precisely and justify your answer.

Case C: A triangle has angles 20, 35, and 125 degrees. Locate its circumcenter relative to the longest side and its opposite vertex, and justify your answer.
1. Hint 1
  Classify the largest angle in each case.
2. Hint 2
  Use the midpoint-parallel argument for the side opposite that largest angle.
3. Hint 3
  In an acute triangle, repeat the same-side conclusion for all three side lines to establish interior location.
Answer

Case A: inside. Case B: at the hypotenuse midpoint. Case C: outside, across the longest side from the 125 degree vertex.

Full solution

Case A: Every angle is acute.
For any side, the midpoint-parallel proof puts the bisector intersection in the half-plane containing the opposite vertex.
The center lies in all three interior half-planes and therefore inside the triangle.

Case B: The side opposite the 90 degree angle is the hypotenuse.
In the midpoint triangle, the connecting midsegment is already perpendicular to the other chosen side, making the hypotenuse midpoint the two bisectors’ intersection.

Case C: Name the side opposite 125 degrees AB and its opposite vertex C.
Let M be the midpoint of AB and N the midpoint of AC.
The midsegment MN is parallel to BC, so angle ANM is also 125 degrees.

The perpendicular at N meets the perpendicular through M across AB from C.
An intersection on C's side would require angle MNO to have the negative measure of 90 minus 125 degrees, as in the lesson's midpoint-parallel argument.
Thus the circumcenter is outside.

The three supplied angle sets give different circumcenter locations. Each dashed circle passes through its triangle’s vertices.

Answer

Case A: inside. Case B: at the hypotenuse midpoint. Case C: outside, across the longest side from the 125 degree vertex.

Key idea

For inside location, check all three interior half-planes, not just one.
Problem 8 Test the distinct-endpoint hypothesis

The perpendicular-bisector locus theorem assumes distinct endpoints A and B. Remove that assumption by taking A and B to be the same point.

Describe all points P with equal distances PA and PB. Explain why this is a counterexample to extending the line-locus conclusion to coincident endpoints.
1. Hint 1
  Compare the two distances when both endpoints have the same location.
2. Hint 2
  Choose any line through the shared endpoint. Compare PA with PB first for a point P on that line, then for a point P away from that line.
Answer

For coincident endpoints, every point of the plane has equal distances to A and B. The locus is the whole plane, not one line.

Full solution

If A and B coincide, PA and PB name the distance from P to the same location.
They are equal for every point P.

For example, P at the shared endpoint has both distances zero.
Points 1 cm away in any direction also have equal distances, as do all other points.

The whole plane satisfies the condition and cannot be described by one perpendicular bisector line.
This demonstrates why the theorem requires distinct endpoints.

Answer

For coincident endpoints, every point of the plane has equal distances to A and B. The locus is the whole plane, not one line.

Key idea

A theorem’s hypothesis controls which locus its conclusion can describe.
Problem 9 Construct a perpendicular from an off-line point

P is off a given line l. Construct the perpendicular from P to l using a compass and unmarked straightedge.

Label the perpendicular foot H. Explain how to ensure that each pair of needed intersections is distinct, and justify the final line.

Construct a perpendicular from the specified point to the given line.

Text description of this figure

Point P is off the given line l. No line through P, foot, or construction crossings are supplied.
1. Hint 1
  Create two points of l equally far from P.
2. Hint 2
  Choose a circle centered at P large enough to cross l twice, at A and B.
3. Hint 3
  Use equal circles centered at A and B with radius AP; besides P, they have a second crossing X.
Answer

Choose a circle centered at P that cuts l at distinct points A and B. Circles centered at A and B with radius AP meet at P and a second point X. Draw PX and mark its intersection with l as H.

Full solution

Increase the radius of a circle centered at P until it crosses l in two distinct points A and B.
Equal radii give equal distances PA and PB.

Triangle APB is nondegenerate.
The strict triangle inequality gives
$AB<2AP.$
Therefore circles centered at A and B with radius AP overlap in two distinct points.

One crossing is P; the other, X, is across l from P.
Both crossings are equally far from A and B, so both lie on their perpendicular bisector.

The joining line PX is that bisector and is perpendicular to l.
Its intersection H with l is the foot.
Mark a right angle at H.

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

Answer

Choose a circle centered at P that cuts l at distinct points A and B. Circles centered at A and B with radius AP meet at P and a second point X. Draw PX and mark its intersection with l as H.

Key idea

Two distinct equal-distance points determine the perpendicular line.
Problem 10 Transfer the theorem to a hidden rotation center

One given 90 degree counterclockwise rotation maps A to A prime and B to B prime. Both point-image chords have nonzero length and are not parallel. Construct its center from the four points on paper, and justify why it is the given rotation center.

Explain how you would repair the choice of pairs if a chord had zero length or both chord bisectors coincided.

Finally, state the additional condition needed if the pairs were arbitrary and were not known to come from one rotation.

Both point-image pairs belong to the same given rotation.

Text description of this figure

A maps to A prime and B maps to B prime in one given 90 degree counterclockwise rotation. The point-image chords are nonzero and nonparallel.
1. Hint 1
  The center of a given rotation has equal distances to each point and its own image.
2. Hint 2
  Apply the converse separately to the two point-image chords.
3. Hint 3
  Explain why nonparallel chords give one bisector intersection, then distinguish center-to-point distance from turn angle.
Answer

Construct the two chord perpendicular bisectors and use their unique intersection O. Replace a fixed-point pair or a pair with the same bisector by another suitable pair. Arbitrary pairs must also require the same directed rotation angle.

Full solution

Construct the perpendicular bisector of AA prime and then that of BB prime.
Their nonparallel chords give nonparallel bisectors, which meet at one point O.

The given rotation center is equally far from A and A prime, so it lies on the first bisector.
It is equally far from B and B prime, so it lies on the second.
Their unique common point O must be that center.

A zero chord comes from a fixed point and has no distinct endpoints.
Replace it with a moved point and its image.
If the bisectors coincide, replace one pair with another whose bisector is different.

For arbitrary pairs, equal radii only put each image on a possible circular path.
The two mappings must also have the same directed turn angle to be one rotation about O.

Each bisector contains the already-given rotation center.

Answer

Construct the two chord perpendicular bisectors and use their unique intersection O. Replace a fixed-point pair or a pair with the same bisector by another suitable pair. Arbitrary pairs must also require the same directed rotation angle.

Key idea

Use the shared rotation hypothesis to recover its center; equal radii alone do not prove one shared angle.

← Back to lesson