The Angle at the Centre

Imagine you and a friend are both watching the same doorway across a room. You stand right in the middle of the room; your friend leans against the far wall. You are both looking at the same doorway — the same two edges — yet because you are standing in different places, the doorway fills a different amount of your view. From the middle of the room it looks wide; from the far wall it looks narrow.

A circle does exactly this, and it does it with a rule so clean it feels almost magical. Pick an arc (or chord) AB on a circle. It is "seen" from two places: from the centre O, and from any point P out on the rim. The centre always sees the arc at exactly twice the angle the rim point does:

\angle AOB = 2 \times \angle APB

Not "about double". Not "double if you draw it neatly". Exactly double — always, for every circle, every arc, every point on the rim. This is the master circle theorem: most of the other circle-angle facts you will ever meet fall straight out of this one.

For an arc (or chord) AB of a circle with centre O:

the arc subtends an angle at the centre \angle AOB, and an angle at the circumference \angle APB at any point P on the circle (on the major arc);
the angle at the centre is exactly double the angle at the circumference — \angle AOB = 2\,\angle APB;
this is the master circle theorem: angles in the same segment, the angle in a semicircle, and more all come straight from it.

Why it works

It rests on one fact you already have: every radius is the same length, so the triangles inside are isosceles. Their equal base angles, added through the exterior-angle rule, give the doubling. Step through it.

In symbols: in isosceles \triangle OAP the base angles are a, and the exterior angle at O is 2a; in \triangle OBP they are b, exterior angle 2b. So \angle AOB = 2a + 2b = 2(a + b) = 2\,\angle APB.

Using it: double or halve

Every problem with this theorem is really the same tiny move. You are given one of the two angles and you want the other:

Going from the rim to the centre? The centre is bigger — so double.
Going from the centre to the rim? The rim is smaller — so halve.

Example 1 — double it. A chord AB subtends an angle of 35^\circ at a point P on the rim. What angle does it make at the centre?

Both angles stand on the same chord, and P is on the far (major) arc, so the theorem applies directly: \angle AOB = 2 \times \angle APB = 2 \times 35^\circ = 70^\circ.

Example 2 — halve it. The angle at the centre on an arc is 140^\circ. What is the angle at the circumference on the same arc?

Now we run the rule the other way — the rim angle is half the centre angle: \angle APB = \tfrac{1}{2}\,\angle AOB = \tfrac{1}{2}\times 140^\circ = 70^\circ. Notice both examples landed on 70^\circ — that is the whole theorem in a nutshell: the two views differ by a factor of two.

Example 3 — a short chase. An arc gives an angle at the centre of 110^\circ, and the triangle formed at the rim point happens to be isosceles with that rim angle as its apex. Find each base angle.

Halve the centre angle for the rim angle: \angle APB = 110^\circ \div 2 = 55^\circ.
That rim angle is the apex of an isosceles triangle, so the two base angles share what's left: (180^\circ - 55^\circ) \div 2 = 125^\circ \div 2 = 62.5^\circ each.

Three lines, one theorem, plus a fact you already knew about isosceles triangles. That "chase a few angles in a row" style is exactly how circle problems are built.

One theorem, a whole toolkit

Here is the beautiful part. Several famous circle theorems are not separate rules to memorise — they are just this one theorem with the numbers filled in.

The angle in a semicircle is 90^\circ. If AB is a diameter, the "arc" is a straight line through the centre, so the angle at the centre is a straight 180^\circ. Half of that is 90^\circ — the angle at the rim is a right angle, every time. That is the angle in a semicircle.
Angles in the same segment are equal. Move the rim point anywhere along the major arc. The centre angle hasn't changed, so its half hasn't either — every rim point on that arc sees the chord at the same angle. That is angles in the same segment.
Opposite angles of a cyclic quadrilateral add to 180^\circ. The two rim points sit on opposite arcs, whose two centre angles fill the whole 360^\circ turn; halving each and adding gives 180^\circ. That is cyclic quadrilaterals.

So you don't really have four theorems to learn. You have one, and three shortcuts that remember its most useful cases.

The "double" only works when everything is lined up correctly. Three traps catch people out:

Same arc, same chord. The two angles must stand on the same arc or chord AB. If the rim angle is looking at a different chord, the theorem says nothing about it.
The rim point must be on the far (major) arc. If P slips onto the near arc, the relevant centre angle is the reflex one (the big one going the long way round). It is still double — but of the reflex angle, so watch which angle you mean. A full turn is 360^\circ, so the reflex angle is 360^\circ minus the ordinary one.
Double one way, halve the other. Centre = 2 \times rim. So to find the centre you double; to find the rim you halve. Students constantly reverse this and double the centre angle by mistake — always ask yourself "which one should be bigger?" The centre, always.

In an old stone archway there is one special wedge at the very top — the keystone. Take it out and the whole arch collapses; leave it in and every other stone leans on it. The angle-at-the-centre theorem is the keystone of circle geometry.

The angle in a semicircle, equal angles in the same segment, supplementary angles in a cyclic quadrilateral — they all lean on this single idea that the centre sees double. It is one of the loveliest things in school maths: not a pile of separate facts to cram, but one deep idea that unfolds, almost on its own, into an entire toolkit. Learn this one properly and the rest come along for free.