Chapter 12.6 — Exercise 12.5 — Cyclic Quadrilaterals

Reading x and y Off Three Figures

All three figures reduce to the identical single equation, "these two opposite angles add to 180°," rearranged around whichever two values are already known. The third figure is the only one needing an extra fact folded in first — recognising that one of its angles sits in a semicircle, and is therefore forced to be exactly 90° before the supplementary-angles equation can even be set up — while the first two are pure substitution the moment the opposite pairs are correctly identified. Identifying which two angles are actually "opposite" in a given figure is itself worth a second look before writing any equation down: opposite means diagonally across the quadrilateral from each other, not simply the two angles that happen to be drawn nearest to each other on the page, and a figure drawn at an unfamiliar angle can make that distinction genuinely easy to misread on a first glance.

The Converse: Supplementary Angles Pull the Fourth Point In

A, B, C already lie together on a circle, and ∠A + ∠C = 180°; prove D also lies on that exact same circle. Since ABCD's four angles must sum to 360° in total, (∠A + ∠C) + (∠B + ∠D) = 360°, so 180° + (∠B + ∠D) = 360° exactly, giving ∠B + ∠D = 180° too, as required. Both pairs of opposite angles are now fully supplementary, which is exactly the defining condition that makes ABCD a cyclic quadrilateral — so D genuinely lies on that exact same circle as A, B, and C.

This is a genuinely useful way to place a fourth point precisely: rather than measuring a distance from some centre, confirming that one pair of opposite angles sums to 180° is enough on its own to guarantee the entire quadrilateral is cyclic, with the second supplementary pair following automatically rather than needing to be checked separately. This is really the same technique already used constantly throughout this chapter to place a point exactly: earlier exercises located a fourth point using a distance condition (equal to a known radius) or a perpendicularity condition (bisecting a chord), and this converse simply adds a third route — an angle-sum condition — to that same growing toolkit of ways to pin a point down without measuring it directly.

A Cyclic Parallelogram Can Only Be a Rectangle

A parallelogram's opposite angles are always already equal (∠A = ∠C); a cyclic quadrilateral's opposite angles are supplementary (∠A + ∠C = 180°). Carefully combining both of these facts about the same single parallelogram: ∠A = ∠C and ∠A + ∠C = 180° together force ∠A = ∠C = 90°, and the same reasoning gives ∠B = ∠D = 90° too. Every angle is a right angle, so the parallelogram is necessarily a rectangle, with no other possibility remaining open.

Two ordinary facts collide here to force something specific: "opposite angles equal" belongs to every parallelogram regardless of whether it's cyclic, and "opposite angles supplementary" belongs to every cyclic quadrilateral regardless of whether it's a parallelogram. A shape satisfying both conditions simultaneously has no freedom left remaining in its angles at all — 90° is the only value solving both equations together, which is exactly why a cyclic parallelogram can never be some other, more slanted shape.

A Cyclic Rhombus Can Only Be a Square

The identical argument applies to a rhombus: its own opposite angles are already equal purely by virtue of it being a parallelogram in the first place, so that same exact combination forces ∠A = ∠C = ∠B = ∠D = 90° the moment cyclic is added into the mix. A rhombus with four right angles added on top of its four already-equal sides is precisely, and only ever, a square.

This result and the rectangle result above are really the same proof run twice on two different starting shapes: cyclic plus parallelogram forces right angles regardless of which extra property (equal diagonals for a rectangle, equal sides for a rhombus) the parallelogram already happened to carry in, which is exactly why a shape satisfying every one of parallelogram, rhombus, rectangle, and cyclic simultaneously can only ever be a square — nothing less specific survives all four conditions imposed together. This is a genuinely useful way to think about the entire family-tree structure introduced back in Chapter 8: adding "cyclic" as one further condition on top of an already-named shape doesn't create some brand-new category needing its own separate theory — it simply narrows down which shapes already in that family tree are still allowed to remain, sometimes all the way down to a single, fully determined shape.

Which Shapes Can Be Inscribed At All

The one genuinely impossible case on this list, the non-rectangular parallelogram, isn't impossible by coincidence — it's the direct, entirely inevitable consequence of the very same proof already given above: the moment a parallelogram is forced to be cyclic, it's simultaneously forced to be a rectangle, so "non-rectangular" and "cyclic parallelogram" describe two conditions that can never be satisfied by the same shape at once. Every triangle, by contrast, is always inscribable no matter its angles, since the circumcircle construction from Exercise 12.3 works identically for acute, right, and obtuse triangles alike — a fact worth remembering as the sharp dividing line between triangles, always inscribable no matter their shape, and quadrilaterals, only ever inscribable once they satisfy the real, genuine angle restriction proved throughout this exercise.

From Circles to Compass and Straightedge

Every theorem across this entire chapter proved a relationship already present in a given figure. Chapter 13, Geometrical Constructions shifts to building figures from scratch instead — using nothing but a compass and straightedge alone to construct angles and shapes meeting exact given conditions, the practical, hands-on counterpart to the proofs this whole chapter has spent its time carefully establishing.