Angle Properties

Draw a circle with centre $O$, mark $A$ and $B$ on it, and mark $P$ somewhere on the major arc. Join $OA$, $OB$, $PA$, $PB$ and $PO$. You now have two triangles sharing the vertex $O$. What kind of triangles are they, and why must they be that kind? That single answer is the whole proof.

One theorem does the work: the angle at the centre is twice the angle at the circumference standing on the same arc. It is proved by joining $PO$ and extending it, which splits the figure into two isosceles triangles, and the exterior-angle result finishes it.

Given: a circle with centre $O$, points $A$, $B$ and $P$ on the circumference.
To prove: $\angle AOB = 2 \times \angle APB$.
Construction: join $PO$ and extend it to a point $Q$ outside the circle.

Consider $\triangle OPA$ first. Since $OP = OA$ (radii), it is isosceles, so $\angle OPA = \angle OAP$, and call each $a$. The exterior angle $\angle AOQ$ equals the sum of the two interior opposite angles, so $\angle AOQ = 2a$.

The same argument on $\triangle OPB$ gives $\angle BOQ = 2b$, where $b = \angle OPB$.

Adding, $\angle AOB = 2a + 2b = 2(a + b) = 2 \times \angle APB$.

Let $P$ and $Q$ both lie on the major arc $AB$. Both $\angle APB$ and $\angle AQB$ stand on the minor arc $AB$, so the main theorem applies to each:

$$\angle AOB = 2\angle APB \quad \text{and} \quad \angle AOB = 2\angle AQB$$

The left side is the same angle in both, so $2\angle APB = 2\angle AQB$ and therefore $\angle APB = \angle AQB$.

That is the whole proof: two things equal to half the same thing are equal to each other. The result is usually stated as "angles in the same segment are equal", and "same segment" is just a way of saying the vertices are on the same side of the chord, hence standing on the same arc.

Let $AB$ be a diameter, so $A$, $O$ and $B$ are collinear. Then $\angle AOB$ is a straight angle:

$$\angle AOB = 180°$$

For any point $P$ on the circle, $\angle APB$ stands on the arc $AB$ not containing $P$, and the main theorem gives

$$\angle APB = \frac{1}{2} \times 180° = 90°$$

So the angle in a semicircle is a right angle, for every position of $P$. Its converse is also true and often more useful in reverse: if $\angle APB = 90°$ then $AB$ must be a diameter, which is how a right angle in a circle diagram tells you where the centre is.

Let $APBQ$ be a cyclic quadrilateral, with $P$ on the major arc $AB$ and $Q$ on the minor arc.

$\angle APB$ stands on the minor arc, so it is half the non-reflex $\angle AOB$. $\angle AQB$ stands on the major arc, so it is half the reflex $\angle AOB$. The two central angles together sweep the full turn:

$$\angle AOB + \text{reflex } \angle AOB = 360°$$

Halving both sides gives $\angle APB + \angle AQB = 180°$. So the opposite angles of a cyclic quadrilateral are supplementary.

A second result follows immediately. The exterior angle at any vertex is supplementary to the interior angle beside it, and the interior angle is supplementary to the opposite one, so the exterior angle of a cyclic quadrilateral equals the interior opposite angle.

Four quick problems. Work each one, then reveal the answer.

1. 1 $P$ is on the major arc and $\angle APB = 40°$. Find the non-reflex $\angle AOB$.

   Show answer
   Both stand on the minor arc, so double it.$80°$

2. 2 $AB$ is a diameter and $P$ is on the circle. Find $\angle APB$.

   Show answer
   The central angle is a straight angle.$90°$

3. 3 In a cyclic quadrilateral, one angle is $105°$. Find the opposite angle.

   Show answer
   Opposite angles are supplementary.$75°$

4. 4 $Q$ is on the minor arc and the non-reflex $\angle AOB = 140°$. Find $\angle AQB$.

   Show answer
   Use the reflex angle, $360° - 140° = 220°$, then halve.$110°$