The converse and the exterior angle

The exterior angle

Extend one side of a quadrilateral past a corner. The angle between the extension and the next side is an exterior angle.

In the picture, the side $AB$ is extended to $E$. The exterior angle at $B$ is $\angle CBE$. The angle $\angle ADC$ is the interior opposite angle. It is the inside angle at the corner across from $B$.

In the picture, $\angle CBE = 78^\circ$ and $\angle ADC = 78^\circ$. They are equal.

Theorem. An exterior angle of a cyclic quadrilateral is equal to the interior opposite angle.

Why it is true

1. $A$, $B$ and $E$ lie on a straight line, so $\angle ABC + \angle CBE = 180^\circ$.
2. $ABCD$ is cyclic, so $\angle ABC + \angle ADC = 180^\circ$.
3. Both sums equal $180^\circ$. So $\angle CBE = \angle ADC$.

In the picture, $\angle ABC = 102^\circ$, and $102 + 78 = 180$.

Example 1. The side $AB$ of a cyclic quadrilateral $ABCD$ is extended to $E$. $\angle CBE = 74^\circ$ and $\angle BAD = 89^\circ$. Find $\angle ABC$, $\angle ADC$ and $\angle BCD$.

1. $\angle ABC = 180^\circ - 74^\circ = 106^\circ$.
2. $\angle ADC = \angle CBE = 74^\circ$.
3. $\angle BCD$ is opposite to $\angle BAD$. So $\angle BCD = 180^\circ - 89^\circ = 91^\circ$.

Check. $89 + 106 + 91 + 74 = 360$.

The converse

The theorem says: if the four corners lie on a circle, then opposite angles add to $180^\circ$. Does it work backwards?

Converse. If the sum of a pair of opposite angles of a quadrilateral is $180^\circ$, then the quadrilateral is cyclic.

A quadrilateral with a pair of supplementary opposite angles can always be drawn inside a circle. We prove this by contradiction. We assume the opposite and show that it cannot be.

Proof of the converse

Let $ABCD$ be a quadrilateral with $\angle B + \angle D = 180^\circ$. Draw the circle through $A$, $B$ and $C$. We claim that $D$ is on it.

Suppose $D$ is not on the circle. Say $D$ is inside it, as in the picture.

1. Extend $AD$ until it meets the circle at $D'$. Join $D'C$.
2. $ABCD'$ is cyclic. So $\angle AD'C = 180^\circ - \angle B$.
3. We are told that $\angle ADC = 180^\circ - \angle B$. So $\angle AD'C = \angle ADC$.
4. But $\angle ADC$ is an exterior angle of the triangle $DD'C$. An exterior angle of a triangle is the sum of the two interior opposite angles. So $\angle ADC = \angle DD'C + \angle DCD'$, and that is bigger than $\angle AD'C$.
5. Step 3 says they are equal, and step 4 says one is bigger. This cannot be.

So $D$ is not inside the circle. In the same way, $D$ cannot be outside it. Hence $D$ lies on the circle, and $ABCD$ is cyclic.

Three ways to show that points are concyclic

For four points $A$, $B$, $C$, $D$ that make a quadrilateral $ABCD$, any one of these is enough.

1. A pair of opposite angles adds to $180^\circ$. For example, $\angle B + \angle D = 180^\circ$.
2. An exterior angle equals the interior opposite angle. For example, $\angle CBE = \angle ADC$.
3. Equal angles on one side of a line. If $C$ and $D$ lie on the same side of $AB$ and $\angle ACB = \angle ADB$, then $A$, $B$, $C$, $D$ lie on one circle. This is the converse of the fact in the first course, that angles in the same part of the circle are equal.

Example 2. In the quadrilateral $PQRS$, $\angle P = 112^\circ$ and $\angle R = 68^\circ$. Is it cyclic?

$112 + 68 = 180$, so the opposite angles $P$ and $R$ are supplementary. By the converse, $PQRS$ is cyclic.

Example 3. In the quadrilateral $PQRS$, $\angle P = 108^\circ$ and $\angle R = 64^\circ$. Can it be cyclic?

$108 + 64 = 172$, not $180$. So it is not cyclic.

Practise with new numbers

Try it in Python

To test four points, find the circle through three of them. Then see whether the fourth is at the same distance from its centre.

Common mistakes

• Pairing the exterior angle with the wrong angle. The exterior angle at $B$ equals the interior angle at the opposite corner $D$, not the one at $A$ or $C$.
• Testing only one angle. One pair of opposite angles adding to $180^\circ$ is enough. But the two angles must be opposite, not neighbours.
• Using the converse on a figure that is not a quadrilateral. The four points must be joined in order round the figure.
• Saying "equal" when you mean "supplementary". Opposite angles of a cyclic quadrilateral add to $180^\circ$. They are not equal, except in a rectangle.

Remember these

• An exterior angle of a cyclic quadrilateral equals the interior opposite angle.
• If a pair of opposite angles of a quadrilateral adds to $180^\circ$, the quadrilateral is cyclic.
• The converse is proved by contradiction, using the exterior angle of a triangle.
• Four points are concyclic if opposite angles are supplementary, or an exterior angle equals the interior opposite angle.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. The side $DC$ of a cyclic quadrilateral $ABCD$ is extended to $F$. If $\angle ADC = 99^\circ$, what is $\angle ABC$?

2. In a cyclic quadrilateral $ABCD$, the side $BC$ is extended to $G$ and $\angle DCG = 62^\circ$. Which interior angle does $\angle DCG$ equal, and how big is it?

3. A quadrilateral has angles $\angle A = 77^\circ$, $\angle B = 104^\circ$, $\angle C = 103^\circ$ and $\angle D = 76^\circ$. Is it cyclic?

4. Why must $\angle A + \angle C = 180^\circ$ once $\angle B + \angle D = 180^\circ$ in a quadrilateral?

5. In the proof of the converse, which fact about triangles tells us that $\angle ADC$ is bigger than $\angle AD'C$?