One rule does most of the work: an inscribed angle is half its arc.
Find the angle
In a circle with centre O, the central angle ∠AOB = 130°. Point C lies on the major arc. What is ∠ACB?
The central angle equals its arc, so arc AB = 130°. ∠ACB is an inscribed angle standing on that same arc, so it is half of it: 65°. C forgets to halve, which is the single commonest error in this topic. A subtracts from 180 as though ACB were a straight line, and D is the reflex angle at the centre.
| Angle | Vertex | Size |
|---|---|---|
| Central | at the centre | equal to its arc |
| Inscribed | on the circle | half its arc |
So a central and an inscribed angle standing on the same arc are always in the ratio 2 : 1. Everything in this lesson follows from that.
Arcs around a whole circle add up to 360°.
∠AOB = 130°
so minor arc AB = 130°
360 − 130 = 230°
the major arc
∠ACB = 130 ÷ 2 = 65°
C is on the major arc, so it looks at the minor one
An inscribed angle halves the arc it faces — the one on the far side, not the one its vertex is sitting on. With C on the major arc, ∠ACB opens onto the minor arc.
Move the vertex anywhere along the arc and the angle does not change.
All three stand on the same chord AB from the same side, so all three are half of arc AB. This is why the rule is often stated as "angles in the same segment are equal".
Because the arc is 180°, and half of 180 is 90.
AB is a diameter
so arc AB = 180°
∠ACB = 180 ÷ 2
inscribed, so halved
= 90°
for every C on the circle
This is Thales' theorem, and it is not a special case to memorise separately — it is the inscribed angle rule applied to an arc that happens to be a semicircle. Spotting a diameter in a diagram usually means a right angle is available.
ABCD is cyclic, ∠A = 85°
all four vertices on the circle
∠C = 180 − 85
opposite angles
= 95°
and ∠B + ∠D = 180 too
Why: ∠A and ∠C stand on the two arcs that together make the whole circle, so their halves add to half of 360°, which is 180°.
The same 2 : 1 rule, used when the unknown is the arc, an x, or an angle inside the circle.
Backwards: an inscribed angle is 27°. What is its arc?
arc = 2 × 27° = 54°
undo the halving: double
With algebra: central ∠AOB = (2y + 10)° and inscribed ∠ACB = 40° stand on the same arc. Find y.
central = 2 × inscribed
same arc, ratio 2 : 1
2y + 10 = 2 × 40 = 80
write the equation
y = 35
check: 2(35) + 10 = 80 = 2 × 40 ✓
Writing 2y + 10 = 40 treats the central and inscribed angles as equal, and gives y = 15 — a central angle of 40°, which would make the inscribed angle 20°, not 40°. Decide which angle is the double one first.
Two chords cross inside the circle, cutting off arcs of 96° and 44° (the arcs between their ends, opposite each other). Find the angle between the chords.
angle = (96° + 44°) ÷ 2
half the SUM of the two arcs
= 70°
between 44° and 96°, as it must be
Why: join two chord ends to make a triangle; the angle between the chords is an exterior angle, equal to the sum of two inscribed angles, 48° + 22° = 70°.
The same rules; each case hides the arc we need.
The angle stands on the major arc: ∠AOB = 100°, and D lies on the minor arc AB. Find ∠ADB.
major arc AB = 360° − 100° = 260°
D faces the far side
∠ADB = 260 ÷ 2 = 130°
an obtuse inscribed angle
Different: a vertex on the minor arc opens onto the major arc, so the angle is more than 90°.
A cyclic quadrilateral with x: ∠A = (2x + 10)° and ∠C = (3x − 5)° are opposite.
(2x + 10) + (3x − 5) = 180
opposite angles are supplementary
5x + 5 = 180, x = 35
solve
∠A = 80°, ∠C = 100°
check: 80 + 100 = 180 ✓
Different: finding x is not the end — the question usually wants the angles.
A triangle of radii: O is the centre, ∠OAB = 32°. Find the inscribed angle ∠ACB on the major arc.
OA = OB, so ∠OBA = 32°
isosceles triangle
∠AOB = 180° − 64° = 116°
angle sum of a triangle
∠ACB = 116 ÷ 2 = 58°
half the central angle
Different: two radii always make an isosceles triangle — that is how the central angle is found when it is not given.
Expect a central, inscribed or cyclic-quadrilateral angle to find, often as an expression in x, and a diameter that hides a right angle. The fastest route is to mark the arc first, then halve (inscribed) or copy (central). The 10-second check: an inscribed angle on a minor arc is less than 90°, and opposite angles of a cyclic quadrilateral add to 180°.
Find the central angle
An inscribed angle ∠PQR measures 38°. What is the central angle ∠POR standing on the same arc?
The inscribed angle is half the arc, so the arc is 2 × 38 = 76°, and the central angle equals its arc: 76°. A halves again instead of doubling — check the direction of the rule before using it. D subtracts from 180, which is the cyclic quadrilateral rule and does not apply here.
Hint: the central angle is the bigger of the two. If your answer is smaller than 38°, you have applied the rule backwards.
