Perpendicular bisector

Take any line segment. Is there a single, well-defined line that passes through its midpoint and is perpendicular to it? Yes , and we can construct it without measuring a single thing.

Idea

A perpendicular bisector of a segment $XY$ is a line that

1. passes through the midpoint of $XY$, and
2. is perpendicular to $XY$.

Every segment has exactly one perpendicular bisector.

Key property. A point $P$ lies on the perpendicular bisector of $XY$ if and only if $PX = PY$. That is, the perpendicular bisector is the set of all points equally far from $X$ and from $Y$.

Why? Suppose $PX = PY$. Drop a perpendicular from $P$ to $XY$ meeting it at $O$. Triangles $POX$ and $POY$ share side $PO$, have equal hypotenuses $PX = PY$, and share a right angle at $O$. By RHS congruence, $OX = OY$ , so $O$ is the midpoint. So every "equidistant" point sits on one specific line: the perpendicular bisector.

The construction (compass + unmarked ruler).

1. Open the compass to any radius bigger than half of $XY$.
2. With centre $X$, draw an arc above and an arc below $XY$.
3. With the same radius and centre $Y$, draw arcs above and below. They meet the previous arcs at two points , call them $A$ (above) and $B$ (below).
4. Draw the line $AB$ with the ruler. This is the perpendicular bisector!

Why it works. By construction, $AX = AY$ (both equal to the chosen radius). So $A$ is equidistant from $X$ and $Y$ , and therefore on the perpendicular bisector. Same for $B$. Two points determine the line, so $AB$ is the perpendicular bisector.

The Śulba-Sūtra version. Ancient Indian geometers did the same thing with a folded rope: fix the ends of a long rope at $X$ and $Y$, find the midpoint of the loose rope, pull it taut above and then below $XY$, mark those positions, and join them. Same construction , different compass!

Worked examples

Example 1. Construct the perpendicular bisector of a $6$ cm segment $PQ$.

Draw $PQ$. Open compass to about $4$ cm (more than half of $6$). Mark arcs above and below from $P$, then from $Q$ using the same radius. Join the two intersection points. The line cuts $PQ$ at its midpoint and at $90°$.

Example 2. Use the perpendicular bisector to find the midpoint of a given segment.

Construct the perpendicular bisector as above. The intersection with the original segment is the midpoint , and this is more accurate than measuring with a ruler.

Example 3. Given a segment $XY$, can two different perpendicular bisectors exist?

No. The set of points equidistant from $X$ and $Y$ is a single line. So the perpendicular bisector is unique.

Example 4. Why must the radius be greater than half the segment?

If the radius is less than half, the arcs from $X$ and $Y$ never meet , no intersection points, no line. If the radius is exactly half, they meet only on the segment itself, not above or below, so you cannot draw a clean line.

Try it yourself

1. Construct the perpendicular bisector of an $8$ cm segment.
2. Draw a $5$ cm segment. Using your construction, mark its midpoint.
3. Why is the perpendicular bisector of $XY$ the set of all points $P$ with $PX = PY$?
4. Construct a segment of length $7$ cm and use the perpendicular bisector to divide it into four equal parts. (Hint: bisect, then bisect each half.)
5. Three towns $A$, $B$ and $C$ want a well that is equally far from $A$ and from $B$. Where could it be?
6. Show that the perpendicular bisector of a chord of a circle passes through the centre.
7. Construct a triangle and the perpendicular bisectors of all three sides. Do they meet at one point? (Yes , the circumcentre.)
8. With an $8$ cm segment, what is the smallest compass radius that will work?

Activity

Take a long piece of string and two pencils. Fix one pencil at each end of a $20$ cm chalk line on the floor. Tie the string between the two pencils so it is taut. Pull the midpoint of the string above the line and mark a point on the floor. Repeat below. Join the two marks. You have just done the Śulba-Sūtra perpendicular-bisector construction!