Chapter 12.4 — Exercise 12.3 — Perpendiculars from Centre

The Perpendicular Bisector Theorem, Both Ways

Both directions matter equally here, just as with the chord-angle theorem before it: drawing the perpendicular first guarantees the bisection, and starting from the bisection instead guarantees the perpendicularity. The proof behind both directions is the same triangle trick used throughout this chapter — OA and OB are equal radii, so △OAM and △OBM are always congruent by RHS (once perpendicularity is given) or SSS (once the midpoint is given), and CPCT delivers whichever half of the theorem wasn't the starting assumption. Notice that RHS is the rule available going one direction (perpendicularity given, right angle plus hypotenuse plus shared side already fixed) while SSS is available going the other (bisection given, all three sides already fixed) — two different congruence rules proving the same underlying relationship, simply because the two directions of the proof start from different known facts about the same figure.

Three Points, Exactly One Circle

Any point on the perpendicular bisector of a segment is equidistant from both of that segment's endpoints. Applied to a triangle's three separate sides all at once, all three perpendicular bisectors meet together at a single shared point — equidistant from all three vertices — which is exactly the centre of the one circle passing through all three. This circle is called the triangle's circumcircle, and its centre is the circumcentre. There is always exactly one circle through any three non-collinear points, never more than one and never zero. "Non-collinear" is doing real work in that statement: if the three points happened to lie on a single straight line instead, their three perpendicular bisectors would all run parallel to each other and never meet at any single point at all, which is exactly why no circle can ever be drawn through three collinear points — the construction that finds a circumcentre for any genuine triangle simply has nothing to intersect once the triangle collapses flat.

The circumcentre's own location is worth noticing as it varies with the triangle's shape: it sits comfortably inside an acute triangle, exactly on the midpoint of the hypotenuse for a right triangle, and genuinely outside the triangle altogether for an obtuse one — the exact same construction method works identically well in every one of these cases, but where the resulting point actually lands depends entirely on the triangle's own angles.

Only two of the three perpendicular bisectors ever genuinely need to be drawn, never all three at once: since any two of them already pin down a single intersection point equidistant from all three vertices, the third perpendicular bisector is guaranteed to pass through that same point automatically, without needing to be drawn or independently checked at all. A related construction problem asks for two circles both passing through the same two points A and B, 5.4 cm apart — solved by drawing a circle of radius 5.4 cm centred at A and a second circle of the same radius centred at B, each one automatically passing through the other's centre precisely because the distance AB was chosen to equal that shared radius.

A Common Chord's Perpendicular Bisector

Two circles, centres M and N, intersect at two points A and B; prove M and N both lie on the perpendicular bisector of common chord AB. Join M to N, M to A, M to B, N to A, N to B, and let O be where MN crosses AB. In △MAN and △MBN: MA = MB (radii of the circle centred at M), NA = NB (radii of the circle centred at N), and MN = MN (shared) — SSS, so △MAN ≅ △MBN, giving ∠AMN = ∠BMN by CPCT.

That angle equality feeds a second congruence: in △MAO and △MBO, MA = MB (radii again), ∠AMO = ∠BMO (just proved), and MO = MO (shared) — SAS, so △MAO ≅ △MBO, giving OA = OB and ∠AOM = ∠BOM by CPCT. Since ∠AOM and ∠BOM form a straight line together, they're supplementary, and being equal too means each is exactly 90°. So MN passes through O, the midpoint of AB, at a right angle — MN is the perpendicular bisector of AB, and both centres lie on it.

This proof chains two separate congruences the way several proofs in earlier chapters did — the first (SSS on the outer triangle) establishes an angle, and the second (SAS on a smaller triangle built from that angle) delivers the actual perpendicularity and bisection together in one final step. Neither congruence alone would be enough on its own — the first proves an angle equality but says nothing yet about O specifically, and only feeding that angle into a second, smaller triangle actually pins down the right angle and the equal segments the problem was originally asking to prove.

Equal Angles at the Intersection Force Equal Chords

Two chords AB and CD of a circle intersect at E, making equal angles with a diameter through E; prove AB = CD. Let OL and OM be the perpendiculars from centre O onto AB and CD. In △OEL and △OEM: ∠OLE = ∠OME = 90° (perpendiculars), ∠OEL = ∠OEM (given), and OE = OE (shared) — AAS, so △OEL ≅ △OEM, giving OL = OM by CPCT. Chords equidistant from the centre are equal in length, so AB = CD.

The perpendicular distance from the centre — OL and OM here — is really the bridge connecting an angle condition to a length conclusion: the given fact is about angles at E, but the conclusion needed is about chord lengths, and "chords equidistant from the centre are equal" is the one theorem in this whole exercise genuinely capable of converting a plain distance equality into a chord-length equality directly, with nothing further needed.

From Distance to Angle Measure

Every theorem in this exercise related a chord's position — its distance from the centre — to its length or its perpendicular bisector. Exercise 12.4 shifts to a different relationship entirely: how exactly the angle a chord's arc subtends at the centre compares to the angle that same arc subtends at any other point sitting on the circle itself.