The perpendicular bisector is the set of points equidistant from two points

Naming, explicitly, the line the last two lessons already leaned on without drawing it.

The last two lessons both used one particular line without ever drawing it: the line a point has to sit on to be exactly as far from B as it is from C. This lesson draws that line, and names what it is.

Two fixed points, B and C, with a line drawn through the midpoint of BC, at right angles to it. A free point, P.

That single line is the boundary between “closer to B” and “closer to C” — everywhere else, one side wins. A line that crosses a segment at its midpoint, at a right angle, is that segment’s perpendicular bisector. The set of every point satisfying some condition — here, “equally far from B and from C” — is called that condition’s locus. This lesson’s claim is that the locus of points equidistant from B and C is exactly the perpendicular bisector of BC: not most of it, not approximately, exactly that line and nothing else.

A claim like that is really two claims fused together, and each needs its own argument.

Every point on the line is equidistant from B and C

Take any point P on the perpendicular bisector, and drop a perpendicular from P to BC — it lands exactly at M, the midpoint, because the line was built through M at a right angle. Triangle PMB and triangle PMC share PM, have equal legs BM and CM by M being the midpoint, and share the right angle at M: SAS, so they are congruent, so PB equals PC.

Every point equidistant from B and C is on the line

This is the first direction’s converse, and — as this course now expects — needs its own argument, not a reversal of the first one for free. If PB equals PC, triangle PBC is isosceles, so the same construction G4.3 used applies: drop a line from P to BC, landing at a point M where the two resulting triangles, PMB and PMC, are forced congruent by SSS, the same way G4.3’s were. That congruence hands over two things this time, not one — BM equals CM, so M is the midpoint, and the two angles at M are equal to each other; since they also sit on a straight line and so must add to 180 degrees, each one is 90 degrees. A midpoint and a right angle together are exactly what “perpendicular bisector” means — so P sits on the line this lesson has been calling that all along.

This is the first time in this course that a claim needed proving in both directions to be complete, not because the second direction was optional extra rigour, but because “the set of points with property X” genuinely means two things: nothing outside the set has the property, and nothing with the property is outside the set. Every locus claim from here on (the angle bisector, next) needs the same two-direction check, so this is the lesson where the habit has to form: state a locus claim, then ask separately whether anything got left in or left out.

P is equidistant from B and C. Is P necessarily the midpoint of BC?

No — the midpoint is only one particular point on the perpendicular bisector. Every other point on that same line is also equidistant from B and C, just not on the segment BC itself.

Two circles, one centred at B and one at C, have the same radius and cross at two points. What can be said about those two crossing points?

Both are equidistant from B and C, by definition of a circle’s radius, so both sit on the perpendicular bisector of BC — this is exactly the compass-and-straightedge construction for finding that line without measuring an angle at all.

Why does this lesson need two separate arguments instead of one?

Because “the locus is the line” is really a claim about a set matching a line exactly, and a set can fail to match a line in two different ways: by including points the line does not, or by leaving out points the line does. Each direction rules out one of those failures, and neither argument, on its own, rules out both.

Do the exercise

This lesson counts as done once you get its exercise right. Nothing to tick off by hand.