# Opposite angles in a cyclic quadrilateral add up to 180°

Categories: gcse geometry

A cyclic quadrilateral is a four-sided shape where each of the four vertices lies on the circumference of a circle, like this:

The opposite angles of a cyclic quadrilateral add up to 180°.

Here is a video:

## Opposite angles in a cyclic quadrilateral

Let's look in a bit more detail at exactly what this means.

In the diagram above, the four points **A**, **B**, **C**, and **D** are all on the circle.

According to the theorem, the opposite angles add up to 180°. There are two pairs of opposite angles. *a* is opposite *c*, and *b* is opposite *d*. So the theorem tells us that:

and

## The converse is also true

It is also true that, if all four vertices do not lie on the circumference of a circle, then the opposite angles do not add up to 180°.

Here is an example:

Here, **P**, **R**, and **S** are all on the same circle, but point **Q** is not. So angles *p* and *r* don't add up to 180°, and neither do angles *q* and *s*.

Notice that there is nothing special about the fact that **P**, **R** and **S** are all on the same circle. It is always possible to draw a circle through any three points. The only time we can't do this would be if **P**, **R** and **S** are all in a straight line, but in that case the shape wouldn't be a quadrilateral.

Having drawn the circle that passes through **P**, **R** and **S**, then for the quadrilateral to be cyclic the point **Q** must also be on that same circle. In this example, **Q** is clearly not on the circle so the quadrilateral is not cyclic.

## Proof

To prove this result, we will rely on the circle theorem the angle at the centre of a circle is twice the angle at the circumference.

This diagram shows points **A**, **B**, and **D** from the original cyclic quadrilateral, together with the angle *a*:

It also shows the centre of the circle, **O**, and the angle at the centre, *x*. The theorem we just mentioned tells us that the angle at the centre (*x*) is twice the angle at the circumference (*a*). So:

This diagram shows points **C**, **B**, and **D** from the original cyclic quadrilateral, together with the angle *c*:

It also shows the angle at the centre, *y*. The same theorem applies and tells that us the angle at the centre (*y*) is twice the angle at the circumference (*c*). So:

As this final diagram shows, angles *x* and *y* make a full turn (360°):

So *x* and *y* add up to 360:

Substituting *2a* for *x* and *2c* for *y* gives:

And dividing both sides by 2 gives:

So we have proven that the opposite angles *a* and *c* add up to 180.

What about the other pair of opposite angles, *b* and *d*? Well, we know that the internal angles of a quadrilateral add up to 360, so:

Sunstituting *a + c* with 180 gives:

So:

This proves that opposite angles *b* and *d* also add up to 180.

## Special case when BD is a diameter

When the line **BD** is a diameter of the circle, it creates a special case. We know that the angle in a semicircle is a right angle, so the triangles **BAD** and **BCD** are right angles triangles. The angles at **A** and **C** are right angles. This is shown in the top left circle below:

This leads to several special cases. If *BA** equals **BC**, the shape forms a right kite. That is a kite where the two opposite angles are 90°. This is shown in the top right circle above.

If the line **AC** is also a diameter, then all four angles are right angles, so the shape is a rectangle. This is shown in the bottom right circle above.

Finally, if the line **AC** is a diameter and it is perpendicular to the diameter **BD**, the shape formed is a square. This is shown in the bottom right circle above.

## Awkard case - quadrilateral doesn't contain the centre O

Here is a case where the quadrilateral doesn't contain the circle centre **O**:

Does the rule apply here? Well yes, it does. Here is a repeat of the proof from earlier, applied to the special case. First, we can see that *x* equals *2a*, as before:

We can also see that *y* equals *2c*:

This case was covered in the angle at the centre of a circle is twice the angle at the circumference mentioned earlier.

Finally, we can see that angle *x* and *y* add up to 360°

These are the three facts that we used to prove the result earlier. These facts are still true in this special case, so the proof still holds.

## See also

- Parts of a circle
- Tangent and radius of a circle meet at 90°
- Two radii form an isosceles triangle
- Perpendicular bisector of a chord
- Angle at the centre of a circle is twice the angle at the circumference
- Angle in a semicircle is 90 degrees
- Angles in the same segment of a circle are equal
- Two tangents from a point have equal length

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