Chapter 12.3 — Exercise 12.2 — Chord and Angle

Equal Chords, Equal Angles — Both Directions

Both directions of this theorem hold true simultaneously, which is exactly what the double arrow symbol means here: knowing the chords are equal proves the angles equal, and knowing the angles are equal proves the chords equal, with neither direction being any more fundamentally "basic" than the other. This is really a direct consequence of SSS congruence: △AOB and △COD always share two radii of equal length (OA = OC and OB = OD, since every radius of the same circle is equal by definition) — so the moment either the third side (the chord) or the angle enclosed between the two known sides is fixed equal, the whole triangle is pinned down by SAS or SSS, forcing everything else about it, including the missing piece, to match too. This is genuinely the same reasoning already used constantly throughout Chapter 7's congruence work, just applied here to a pair of triangles that happen to share their two equal sides for a specific, circle-related reason — because both sides in each triangle are radii of the same circle — rather than because the problem happened to state that fact directly.

Every single triangle formed by joining two points on a circle back to its centre automatically has two equal sides, no matter which two points are chosen — a fact so constant across every problem in this chapter that it rarely gets restated explicitly after the first few times, yet it's doing real, load-bearing work in nearly every proof that follows.

A Direct Substitution

If AB = CD and ∠AOB = 90°, find ∠COD. Since AB = CD, the theorem gives ∠AOB = ∠COD directly — so ∠COD = 90° follows immediately, matching the given angle exactly. Nothing beyond a single substitution is needed here; the entire problem exists just to confirm the theorem's statement can be read and applied immediately, before the next two problems ask for something less direct. Recognising this problem as a pure substitution rather than something needing further reasoning is itself worth noticing: the theorem is stated as an if-and-only-if, so once "AB = CD" is confirmed as given, "∠AOB = ∠COD" is available immediately as an equally valid restatement of the exact same fact, not a separate conclusion still needing its own justification beyond simply invoking the theorem by name.

Chasing Angles Through Two Isosceles Triangles

If PQ = RS and ∠ORS = 48°, find ∠OPQ and ∠ROS. Since OR = OS, both being radii of the same circle, △ORS is isosceles, so its two base angles are equal: ∠OSR = ∠ORS = 48°. Its third angle then follows from the angle sum of any triangle: ∠ROS = 180° − 48° − 48° = 84° exactly. Since PQ = RS is given, the chord-angle theorem immediately gives ∠POQ = ∠ROS = 84° as well. Triangle OPQ is isosceles too, since OP = OQ are both radii of the same circle, so its two base angles must be equal: 2∠OPQ = 180° − 84° = 96°, giving ∠OPQ = 48° as the final answer.

Every angle in this figure ultimately traces back to the one given fact, 48°, through a chain of isosceles triangles rather than any single formula — both △ORS and △OPQ are isosceles for the identical reason, two of their sides being radii of the same circle, which is a fact every triangle drawn from the centre to two circle points shares automatically, whether or not the problem bothers to point it out directly. The two isosceles triangles here also aren't independent of each other, even though they're solved one after the other: △OPQ's own vertex angle, ∠POQ, is exactly what the chord-angle theorem hands over from △ORS's result — meaning the second triangle's entire solution genuinely depends on the first one being finished correctly, rather than the two being two separate, unrelated calculations that simply happen to share the same figure.

Diameters Forcing a Parallelogram

PR and QS are two diameters of a circle; is PQ = RS? Since PR and QS are both diameters, they necessarily cross at the centre O, so OP = OR and OQ = OS, since all four of these segments are simply radii of the same circle. This means the diagonals of quadrilateral PQRS — namely PR and QS — bisect each other at the single shared point O, exactly as required by the parallelogram converse theorem. A quadrilateral whose diagonals bisect each other is a parallelogram, so PQRS is a parallelogram, and opposite sides of any parallelogram are always equal, giving the final result: PQ = RS.

This problem never touches the chord-angle theorem at all — it reaches back instead to the parallelogram converse theorem from Chapter 8 (diagonals bisecting each other implies a parallelogram), proved there using ordinary congruence and reused here directly, without needing any modification for the circle context at all. It's a useful reminder that a circle problem doesn't have to be solved with circle-specific theorems just because a circle happens to be drawn in the figure; sometimes the fastest route runs straight through a fact from an entirely different chapter. Spotting that a figure is secretly a parallelogram, hiding inside what first looks like a purely circular setup, takes a small shift in attention: instead of asking "what do I know about this circle," the more useful question becomes "what quadrilateral is quietly formed by the points already given, and what do I already know about that shape instead." Two diameters crossing at a single shared centre is exactly the kind of setup that keeps producing a parallelogram this way, since the crossing point is guaranteed to bisect both diagonals at once purely by being the centre both diameters pass through.

Where the Perpendicular From the Centre Comes In

Every problem here worked with the full angle a chord subtends at the centre. Exercise 12.3 introduces a related but different tool — the perpendicular dropped from the centre onto a chord — and the surprising amount it reveals about where exactly that chord sits. Where this exercise leaned on the full triangle formed by a chord and two radii, the next one leans instead on cutting that same triangle exactly in half — a small shift in construction that turns out to unlock an entirely different, equally useful family of results about chord lengths and their distance from the centre.