Angles in a semicircle and cyclic quadrilaterals

Finding the centre with the corner of a page

Here is a circle drawn on paper. Where exactly is its centre? You have a pencil, a ruler and an ordinary sheet of paper, but no compasses.

There is a trick that works every time. Put one square corner of the sheet exactly on the edge of the circle. The two edges of the sheet cross the circle at two more points. Mark them, and join them with a straight line. Now turn the sheet, put its corner somewhere else on the circle, and do the same again. The two lines you drew cross at the centre.

It works for any circle and any two places on its edge. But why? Nothing in the trick mentions the centre. The only special thing is the square corner, a right angle. By the end of this lecture you will see that each line you drew is a diameter, and the reason comes straight out of the last lecture's theorem.

The special case the last lecture promised

The last lecture promised to take the special case where A and B are the two ends of a diameter. Here it is.

Draw a circle with centre O, and a diameter AB. Put a point P anywhere else on the circle, and draw the triangle APB. What is $\angle APB$?

Use the angle at the centre theorem. Its partner at the centre is $\angle AOB$, standing on the same arc. But A, O and B lie on one straight line, because a diameter passes through the centre. So the "angle" at O is a straight angle:

The angle at the circumference is half the angle at the centre on the same arc. This means that

That is it. That is the whole proof. A diameter makes the angle at the centre a straight line, and half of a straight line is a right angle.

The proof never needed to know where P is. So P can be anywhere on the circle except A or B. Slide P along the edge and the triangle changes shape completely, but the angle at P stays exactly 90°.

The diameter cuts the circle into two semicircles, and P sits on one of them.

A second proof, with no angle at the centre

Here is a second proof that uses only facts from much earlier. It is the one Euclid gives in his Elements, Book III, Proposition 31.

Join P to the centre O. Now there are two small triangles, OAP and OBP. OA, OB and OP are all radii, so they are all equal. This means both small triangles are isosceles.

In triangle OAP, the two equal sides are OA and OP. So the angles opposite them are equal: $\angle OAP=\angle OPA$. Call each one α. In triangle OBP, OB = OP, so $\angle OBP=\angle OPB$. Call each one β.

Now look at the big triangle APB. Its angle at A is α. Its angle at B is β. Its angle at P is made of two parts, α and β. Its three angles add up to 180°:

This means $2(\alpha +\beta )=180^{\circ }$, so $\alpha +\beta =90^{\circ }$. But $\alpha +\beta$ is exactly the angle at P. So $\angle APB=90^{\circ }$.

The first proof shows where the 90° comes from: half of a straight line. The second shows how the triangle shares its 180°: the angle at P always takes exactly half.

The result is often called Thales' theorem, after Thales of Miletus, who lived about 600 BCE. The credit is uncertain: it rests on a writer living more than 600 years later, and other ancient writers gave it to Pythagoras.

Only a diameter gives a right angle

Students often use this result where it does not belong. Test yourself.

So the diameter is the boundary. Points on the larger arc see a shorter chord at an acute angle, and points on the smaller arc at an obtuse angle. Only a diameter, with two equal arcs, gives 90° from both sides.

The result also works backwards. Suppose a triangle APB, with all three corners on a circle, has a right angle at P. Then the angle at the centre on the same arc is $2\times 90^{\circ }=180^{\circ }$. An angle of 180° is a straight line, so A, O and B are in a line. This means AB passes through the centre: it is a diameter. This backwards result is called the converse.

Now the puzzle at the start has its answer. The square corner of the sheet sits on the circle, so the angle there is 90°. By the converse, the two points where the edges cross the circle are the ends of a diameter. So each line you drew is a diameter, and two different diameters meet only at the centre.

Four corners on one circle

Now move from three points on a circle to four. A cyclic quadrilateral is a quadrilateral whose four corners all lie on one circle. Some books say the four points are "concyclic". We name the corners in order around the circle, so ABCD means you meet A, B, C and D as you go round.

In ABCD, the angles at A and C are opposite angles, and so are those at B and D.

Why opposite angles add up to 180°

Draw a cyclic quadrilateral ABCD with centre O. Call the angle at A x, and the angle at C y. Join O to B and to D.

Step 1. The angle at A has arms AB and AD. It stands on the arc across from A, the arc BCD. The angle at the centre on that arc is double, so it is 2x.

Step 2. The angle at C has arms CB and CD. It stands on the other arc, the arc BAD. The angle at the centre on that arc is 2y. In the figure it is a reflex angle, but the theorem allows that, as the last lecture showed.

Step 3. The two arcs BCD and BAD together make the whole circle. So the two angles at the centre together go all the way round O:

Step 4. Divide both sides by 2:

That is the whole idea. Two opposite corners stand on the two arcs that make up the whole circle. Their angles at the centre share 360°, so the corners themselves share 180°. The same argument with B and D, using the arcs CDA and ABC, gives $\angle B+\angle D=180^{\circ }$. Euclid proves this in Book III, Proposition 22.

Two angles that add up to 180° are called supplementary. So you can also say: opposite angles of a cyclic quadrilateral are supplementary.

This fits with the semicircle. If AC is a diameter, the angles at B and D are both 90°, and $90^{\circ }+90^{\circ }=180^{\circ }$. The two results agree because they come from the same parent.

Which quadrilaterals fit inside a circle?

Every triangle has a circle through its three corners, but most quadrilaterals do not. Draw the circle through A, B and D. If C is inside it, the angle at C is bigger than a point on the circle would see; if C is outside, smaller. Only when C is on the circle does $\angle C$ equal $180^{\circ }-\angle A$. So the converse is true: if one pair of opposite angles adds up to 180°, the quadrilateral is cyclic. This gives a test you can use on any quadrilateral. Use it on the cards below.

Did you see the pattern? A shape with equal opposite angles fits in a circle only when they are 90°.

A shortcut from the same fact: the exterior angle

In the figure, the side DC of a cyclic quadrilateral is extended past C to E. The angle BCE is an exterior angle.

The angles on the straight line DCE add up to 180°, so $\angle BCE=180^{\circ }-\angle BCD$. The theorem says $\angle BAD=180^{\circ }-\angle BCD$ too. Both are "180° minus the same angle", so they are equal:

In words: the exterior angle of a cyclic quadrilateral equals the interior angle at the opposite corner.

Worked examples

In every example, the letters are points on the circle, in order.

Three traps