What is a Perpendicular Bisector
A perpendicular bisector is a line that does two things to a segment at once: it crosses the segment at its midpoint (so it bisects, cutting the segment into two equal halves) and it meets the segment at a right angle (so it is perpendicular). Both conditions must hold — a line through the midpoint at a slant is only a bisector, and a perpendicular line that misses the midpoint is only a perpendicular.
The word bisect means to bisect — to divide into two equal parts — and perpendicular means at 90°. Put together, the perpendicular bisector is the unique line that splits a segment evenly and squarely. Every segment has exactly one.
The Perpendicular Bisector Theorem
The perpendicular bisector theorem states:
Any point lying on the perpendicular bisector of a segment is equidistant from the two endpoints of that segment.
In symbols, if line $\ell$ is the perpendicular bisector of segment $AB$, and $P$ is any point on $\ell$, then $PA = PB$. The word equidistant simply means "the same distance from" — point $P$ is equally far from $A$ as it is from $B$.
The strength of the theorem is the word any. It does not matter where on the perpendicular bisector you stand — at the midpoint, far up the line, far down the line — the two distances to the endpoints stay equal. The whole line is, in effect, a set of points all balanced perfectly between $A$ and $B$.
How Do You know A Point is Equidistant Without Measuring?
You do not have to measure at all. If you can confirm a point lies on the perpendicular bisector, the theorem hands you the equal distances for free. That is the practical value: a single fact about position (on the line) replaces two separate distance measurements.
Proof of the Perpendicular Bisector Theorem
The proof rests on the SAS congruence rule — Side-Angle-Side. Here is the setup and the reasoning, each step on its own line.
Given: Line $\ell$ is the perpendicular bisector of segment $AB$, meeting it at midpoint $M$. Point $P$ is any point on $\ell$.
To prove: $PA = PB$.
Drop the proof by comparing triangle $PMA$ with triangle $PMB$.
Step 1: $AM = MB$, because $M$ is the midpoint of $AB$ (the bisector cuts the segment in half).
Step 2: $\angle PMA = \angle PMB = 90°$, because $\ell$ is perpendicular to $AB$.
Step 3: $PM = PM$, the shared side common to both triangles.
Step 4: By SAS (two sides and the included angle equal), triangle $PMA \cong$ triangle $PMB$.
Step 5: Corresponding parts of congruent triangles are equal, so $PA = PB$.
That completes the proof: any $P$ on the perpendicular bisector satisfies $PA = PB$.
The Converse of the Perpendicular Bisector Theorem
Every theorem has a converse — the statement with its "if" and "then" swapped — and for this theorem the converse is just as useful, and also true.
Converse: If a point is equidistant from the two endpoints of a segment, then that point lies on the perpendicular bisector of the segment.
Read the two together. The theorem says on the line $\Rightarrow$ equidistant. The converse says equidistant $\Rightarrow$ on the line. Because both directions hold, "lying on the perpendicular bisector" and "being equidistant from the endpoints" are two names for the same condition. The converse is what lets you find the perpendicular bisector: locate the points equal in distance from both ends, and you have traced the line.
The converse is proved the same way, again leaning on triangle congruence to show that an equidistant point must sit on the perpendicular through the midpoint.
Where the Theorem Leads: The Circumcentre
The theorem pays off most visibly inside a triangle. Each of a triangle's three sides has its own perpendicular bisector, and all three of those lines meet at a single point — the circumcentre.
Why do they all meet at one point? Take the meeting point of two of the perpendicular bisectors; by the theorem, it is equidistant from the endpoints of each side it bisects, which chains together to make it equidistant from all three vertices. Being equal in distance from every vertex, it must (by the converse) lie on the third perpendicular bisector too. The circumcentre of a triangle is therefore the one point equally far from all three corners, exactly the centre of the circle that passes through all three.
Examples of the Perpendicular Bisector Theorem
The examples progress from a direct distance read-off to setting up and solving an equation from the equal-distance condition.
Example 1
Point P lies on the perpendicular bisector of segment AB. If PA = 9 cm, find PB.
By the perpendicular bisector theorem, any point on the perpendicular bisector is equidistant from the endpoints.
So $PB = PA = 9$ cm.
Final answer: PB = 9 cm.
Example 2
Line ℓ is the perpendicular bisector of CD, meeting it at M. If CD = 16 cm, find CM.
The perpendicular bisector passes through the midpoint, so it cuts CD into two equal halves.
$CM = \frac{CD}{2} = \frac{16}{2} = 8$ cm.
Final answer: CM = 8 cm.
Example 3
A point Q is the same distance from both ends of segment EF. A student concludes Q must be the midpoint of EF. Is that correct?
The tempting wrong path: equating "equidistant from the endpoints" with "the midpoint," since the midpoint is obviously equidistant.
Watch where that breaks. The midpoint is one equidistant point, but it is not the only one. By the converse of the theorem, every point on the perpendicular bisector is equidistant from E and F — and almost all of those points are nowhere near the segment itself.
The correct reading: Q lies on the perpendicular bisector of EF, but it need not be the midpoint. The midpoint is just the single equidistant point that happens to sit on the segment.
Final answer: No. Q lies on the perpendicular bisector of EF, which contains the midpoint but infinitely many other points too.
Example 4
Point P is on the perpendicular bisector of segment AB. PA = (3x + 5) and PB = (5x − 7). Find x.
The theorem guarantees $PA = PB$, so set the two expressions equal.
$3x + 5 = 5x - 7$
$5 + 7 = 5x - 3x$
$12 = 2x$
$x = 6$
Final answer: x = 6.
Example 5
In a triangle, point O is the circumcentre. The distance from O to vertex A is 10 cm. What is the distance from O to vertex B?
The circumcentre is equidistant from all three vertices, because it lies on the perpendicular bisector of every side.
So $OB = OA = 10$ cm.
Final answer: OB = 10 cm.
Example 6
Point R lies on the perpendicular bisector of GH. RG = (2y + 4) and RH = (4y − 10). After finding y, give the common distance RG.
Since R is on the perpendicular bisector, $RG = RH$.
$2y + 4 = 4y - 10$
$4 + 10 = 4y - 2y$
$14 = 2y$
$y = 7$
Substitute back to find the distance:
$RG = 2y + 4 = 2(7) + 4 = 18$
Final answer: y = 7 and RG = 18.
Why the Perpendicular Bisector Theorem Matters
The theorem exists because "the set of points equally far from two fixed places" turns out to be one of the most useful objects in geometry — and it is always a straight line, the perpendicular bisector.
That single idea solves real problems:
Locating a fair midpoint. To place a facility, a tower, or a road equally between two towns, you find a point on the perpendicular bisector of the line joining them. Every such point is genuinely equidistant.
Drawing the circumscribed circle. Finding the one circle that passes through three given points means finding their circumcentre — the meeting point of perpendicular bisectors.
Navigation and GPS-style positioning. Pinning a location from equal-distance readings to known points uses the equidistant-line idea directly.
What links these is the theorem's promise: position on the line is equal distance. You never have to measure both distances separately — the line guarantees them.
The Mistakes Students Make Most Often
The errors here come from blurring "perpendicular bisector" with the simpler ideas it is built from. Three are most common.
Mistake 1: Dropping one of the two conditions
Where it slips in: When a line passes through the midpoint but not at a right angle, or is perpendicular but misses the midpoint, and a student still calls it a perpendicular bisector.
Don't do this: Treat a segment bisector alone, or a perpendicular line alone, as a perpendicular bisector.
The correct way: Both conditions are required — through the midpoint and at 90°. Drop either one and the equidistance theorem no longer applies. A common first instinct is to accept any midline as "good enough," which quietly breaks the result.
Mistake 2: Confusing the equidistant point with the midpoint
Where it slips in: When a point is equidistant from the endpoints and a student assumes it must be the midpoint.
Don't do this: Conclude "equidistant from A and B" means "the midpoint of AB."
The correct way: The midpoint is only one of infinitely many equidistant points. By the converse, every point on the perpendicular bisector is equidistant. The midpoint is simply the one that lands on the segment itself.
Mistake 3: Setting the distance expressions equal incorrectly
Where it slips in: In algebra problems where PA and PB are expressions, students sometimes add them or set one to zero instead of equating them.
Don't do this: Write $PA + PB = 0$ or solve only one expression on its own.
The correct way: The theorem gives $PA = PB$, so set the two expressions equal and solve. The error that costs the most marks is forgetting that the theorem's output is an equation between the two distances, not a single value.
Conclusion
The perpendicular bisector theorem says any point on the perpendicular bisector of a segment is equidistant from its two endpoints.
A perpendicular bisector must pass through the midpoint and meet the segment at a right angle — both conditions.
The theorem is proved using SAS congruence of the two triangles formed at the midpoint.
Its converse is also true, so equidistant points are exactly the points on the perpendicular bisector.
The three perpendicular bisectors of a triangle meet at the circumcentre, equidistant from all three vertices.
Because this theorem feeds directly into circle geometry and coordinate proofs, working it through with a teacher builds a foundation that pays off later. Explore Bhanzu's geometry tutor, our middle school math tutor sessions, or math classes online to practise these proofs live.
A Practical Next Step
Work through these problems to solidify your understanding:
If P is on the perpendicular bisector of AB with $PA = (4x - 1)$ and $PB = (2x + 9)$, find $x$ and the common distance.
Explain why the circumcentre is equidistant from all three vertices of a triangle.
If you get stuck on the algebra, return to Example 4. Want your child to build these proof skills with a live trainer? Book a free demo class.
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