Angle subtended by an arc

Let $AB$ be an arc of a circle with centre $O$. The angle $\angle AOB$, measured on the side of the arc under consideration, is the angle subtended by the arc at the centre. If $P$ is any point on the remaining part of the circle — the major arc, when $AB$ is taken as the minor arc — then $\angle APB$ is the angle subtended by the arc at the point $P$, also called the angle in the corresponding segment. The central theorem ties these two angles together in a fixed ratio, no matter where $P$ sits on the remaining arc:

where: $\angle AOB$ is the angle the arc $AB$ subtends at the centre $O$, and $\angle APB$ is the angle the same arc subtends at any point $P$ on the remaining part of the circle. Two immediate ideas follow from this one relationship. First, since every point on the remaining arc gives the same central angle $\angle AOB$, every such point must give the same value of $\angle APB = \tfrac12\angle AOB$ — this is exactly the next NCERT theorem, angles in the same segment are equal, stated without proof. Second, when $AB$ happens to be a diameter, the arc subtends a straight angle $180°$ at the centre, so every point $P$ on the circle sees a right angle:

This is the well-known special case: the angle in a semicircle is always $90°$.

The goal is to prove, for the common case where the centre $O$ lies inside $\angle APB$, that the angle an arc subtends at the centre is exactly double the angle it subtends at any point on the remaining part of the circle, using only the isosceles-triangle property of radii and the exterior-angle property of a triangle.

Step 1. Join $P$ to $O$ and extend the line through $O$ to meet the circle again at $C$, so that $P$, $O$, $C$ lie on one straight line through the centre. This construction splits the figure into two smaller triangles, $\triangle OAP$ and $\triangle OBP$, each of which can be handled with a simple isosceles argument.

Step 2. In $\triangle OAP$, the sides $OA$ and $OP$ are both radii of the circle, so $OA = OP$ and the triangle is isosceles, giving $\angle OAP = \angle OPA$. Applying the exterior-angle property of a triangle — an exterior angle equals the sum of the two interior opposite angles — to the exterior angle $\angle AOC$ gives

Step 3. By the identical argument applied to the isosceles triangle $\triangle OBP$ (since $OB = OP$, both radii), the exterior angle $\angle BOC$ satisfies

Step 4. Add the two results from Steps 2 and 3. Since $\angle AOC + \angle BOC = \angle AOB$ and $\angle OPA + \angle OPB = \angle APB$, the addition gives

which is the required theorem. When $AB$ is instead the major arc and $\angle APB$ lies in the minor segment, the same construction with subtraction in place of addition gives the reflex form of the identical relationship. $\blacksquare$

Problem. In a circle with centre $O$, arc $AB$ subtends an angle of $110°$ at the centre. Find the angle subtended by the same arc at a point $P$ on the major arc, and hence find $\angle ACB$ where $C$ is any other point on the major arc.

Step 1. Apply the theorem directly: the angle at the centre is double the angle at any point on the remaining part of the circle, so

Step 2. Since $C$ is another point on the same major arc as $P$, it lies in the same segment, so by the immediate consequence of the theorem (angles in the same segment are equal), $\angle ACB$ must equal $\angle APB$:

The angle at $P$ is $55°$, and $\angle ACB$ is also $\boxed{55°}$, illustrating that every point on the major arc sees the identical inscribed angle.

Competitive stretch

If instead $AB$ were a diameter (so the central angle were $180°$ rather than $110°$), the same theorem would instantly force $\angle APB = 90°$ for any point $P$ on the circle — the semicircle special case — without needing a separate proof each time.

• Forgetting which arc is being referred to. An arc has a major and a minor part, and each subtends a different angle at the centre; be clear which arc, and which remaining part of the circle, the angle is measured on.
• Doubling in the wrong direction. The centre angle is double the circumference angle, not the reverse — a frequent slip is halving the given centre angle incorrectly or doubling a circumference angle that was already the final answer.
• Assuming any two points give the same angle regardless of side. The equal-angle property only holds for points on the same arc or segment; a point on the other side sees a different, related angle instead.
• Missing the semicircle special case. Diagrams often hide a diameter in plain sight; not spotting it costs an easy $90°$.

Ask the tutor: want to see the proof worked out for the case where the centre lies outside angle APB, or get more practice combining this theorem with the triangle angle-sum property? Ask the AI tutor about "angle subtended by an arc."