Chapter 12.5 — Exercise 12.4 — Arc and Angle

Double at the Centre, Constant Everywhere Else

Moving P anywhere else along the same major arc never changes ∠APB at all — it stays locked at exactly half of ∠AOB no matter where P sits, which is exactly why every point in that same segment subtends an equal angle. Three special cases fall directly out of this doubling relationship, each one just a different value of the central angle: an angle in a semicircle is always 90° (since a diameter's central angle is 180°, and half of that is 90°); an angle in a major segment is always acute (since it corresponds to a central angle under 180°); and an angle in a minor segment is always obtuse, since it now corresponds to a central angle greater than 180° once the reflex angle around the far side of the circle is what's actually being doubled.

Cyclic Quadrilaterals: The Other Half of This Chapter

Points lying on the same circle are concyclic. When all four vertices of a quadrilateral are concyclic, that quadrilateral is cyclic, and its opposite angles are always supplementary — a fact that runs in both directions, just like the chord and perpendicular theorems before it — knowing the quadrilateral is cyclic proves the opposite angles supplementary, and knowing the opposite angles are supplementary proves the quadrilateral cyclic, with neither direction needing the other to already be established first. This converse direction is genuinely useful on its own: given any quadrilateral where three vertices are already known to sit on a circle, showing that its opposite angles sum to 180° is enough to guarantee the fourth vertex lies on that identical circle too, without ever needing to measure a distance from the centre directly.

Chaining Centre-Angle and Cyclic-Quadrilateral Facts

∠AOB = 100° at the centre; find ∠ADB, where D is a fourth point on the circle. First, ∠ACB = ½∠AOB = 50° exactly (centre-angle theorem, using any point C on the major arc). Since ACBD is a cyclic quadrilateral, ∠ACB + ∠ADB = 180°, so ∠ADB = 180° − 50° = 130° in total.

A related problem skips the centre entirely: ∠POR = 120° gives ∠PQR = ½(120°) = 60° the same way, and then ∠PSR = 180° − 60° = 120° by the same cyclic-quadrilateral step. Both problems chain the exact same two facts in the exact same order — halve the central angle to reach the circumference angle, then subtract from 180° to reach the opposite vertex's angle — which is really the two theorems from this exercise working together as a single two-step tool. Neither theorem alone would finish either problem: the centre-angle theorem stops at the circumference angle on the near side, and the cyclic-quadrilateral theorem needs that near-side angle already in hand before it can reach across to the opposite vertex — the two facts are genuinely complementary, each one picking up exactly where the other leaves off.

Angles in the Same Segment Need No Centre At All

∠BAD = 40°; find ∠BCD, where A, B, C, D all lie on one circle. Since ∠BAD and ∠BCD both subtend the same arc BD from the same segment, they're equal directly: ∠BCD = 40°. No centre angle is needed anywhere in this problem — "angles in the same segment are equal" applies directly between two circumference points, without ever routing through O at all, which makes this the fastest possible problem in the whole exercise once the same-segment condition is spotted. Recognising that condition quickly comes down to one visual check: do both angles' vertices sit on the same side of the shared chord, looking at the same arc from the same side? If so, "angles in the same segment" applies immediately and no further construction, centre angle, or cyclic-quadrilateral step is needed anywhere in the solution at all.

Recovering a Radius From a Bisected Chord

OM = 3 cm, AB = 8 cm, OM ⊥ AB; find the radius. Since the perpendicular from the centre always bisects the chord, AM = 4 cm exactly. Triangle OAM is right-angled at M, so by Pythagoras: OA² = OM² + AM² = 9 + 16 = 25, giving OA = 5 cm.

This problem reaches directly back into Exercise 12.3 for its first step (perpendicular bisects chord) before finishing with the Pythagorean theorem — a reminder that this chapter's exercises keep building on each other rather than standing alone, even when a problem is filed under a later exercise's heading. It's worth noticing, too, that this specific right triangle — radius as hypotenuse, half-chord and perpendicular distance as the two legs — recurs constantly throughout circle geometry generally, well beyond this one textbook chapter; any problem pairing a chord's length with its distance from the centre almost always reduces to exactly this same Pythagorean setup.

Two Closing Problems, Two Different Shortcuts

Neither of these final two problems needs a single new circle theorem — both reach back into earlier material (equal chords from Exercise 12.3, equal diagonals of a square from Chapter 8) and simply recognise that the circle context doesn't change how those older facts apply. Spotting that a "circle problem" is secretly a shortcut through an already-proved fact from an earlier chapter is a skill this whole chapter keeps rewarding, not a special trick reserved only for these two particular problems. Both closing results also make a genuinely useful point about circles that don't look identical at first glance: the equal-chord fact and the square's-diagonal fact are two completely different starting observations, yet both land on the exact same kind of answer — a single recovered radius — simply because a radius is, in the end, always just some fixed distance from the centre out to the boundary, however that distance happens to get established.

The Chapter's Final Shape: Special Cyclic Quadrilaterals

Every theorem here connected a circle's centre, its arcs, or its inscribed angles to one another. Exercise 12.5 closes the chapter by asking which familiar quadrilaterals — parallelograms, rhombuses — can ever actually be cyclic, and what extra rigid structure being genuinely cyclic actually forces onto each one of them.