Why the Opposite Corners of a Quadrilateral in a Circle Always Add to 180°

A cyclic quadrilateral is a four-sided figure whose four vertices all lie on one circle. Its sides are chords of that circle, so the circle angle theorems from Part 1 apply to it.

The key result is that **opposite angles always add to $180^{\circ }$.** From it come the exterior angle property, tests for concyclic points, and many multi-step problems.

This part covers each of these in turn.

**In a cyclic quadrilateral $ABCD$, $\angle A+\angle C=180^{\circ }$ and $\angle B+\angle D=180^{\circ }$.

Why.** $\angle A$ and $\angle C$ stand on the two arcs $BCD$ and $BAD$. The angles these arcs subtend at the centre add to $360^{\circ }$, and each angle at the circumference is half of its centre angle, so the two add to $180^{\circ }$.

Worked example 1. $\angle A=75^{\circ }$, so $\angle C=105^{\circ }$.

Worked example 2. $\angle B=(2x+10{)}^{\circ }$ and $\angle D=(3x-5{)}^{\circ }$.

$$2x+10+3x-5=1805x=175x=35\angle B=80^{\circ },\angle D=100^{\circ }$$

Worked example 3. $O$ is the centre and $\angle BOD=140^{\circ }$, with $A$ on the major arc. Then $\angle BAD=70^{\circ }$ and $\angle BCD=180^{\circ }-70^{\circ }=110^{\circ }$.

An everyday example. Four diyas placed on the rim of a circular rangoli form a cyclic quadrilateral, and the angles at opposite diyas always pair up to $180^{\circ }$.

The substance. A parallelogram can be cyclic only if it is a rectangle, since its opposite angles are equal and must also add to $180^{\circ }$.

If one side of a cyclic quadrilateral is produced, the exterior angle formed equals the interior angle at the opposite vertex.

Why. Produce $BC$ to $E$. Then $\angle DCE=180^{\circ }-\angle BCD$, and $\angle BAD=180^{\circ }-\angle BCD$ because opposite angles are supplementary. So $\angle DCE=\angle BAD$.

Worked example 1. $\angle BAD=85^{\circ }$ and $BC$ is produced to $E$. Then $\angle DCE=85^{\circ }$.

Worked example 2. $AB$ is produced to $E$ and $\angle CBE=72^{\circ }$. The opposite interior angle is at $D$, so $\angle ADC=72^{\circ }$ and $\angle ABC=180^{\circ }-72^{\circ }=108^{\circ }$.

Worked example 3. In cyclic $ABCD$, $DC$ is produced to $F$, $\angle BCF=95^{\circ }$ and $\angle ABC=70^{\circ }$.

$$\angle BAD=95^{\circ }\angle ADC=180^{\circ }-70^{\circ }=110^{\circ }\angle BCD=180^{\circ }-95^{\circ }=85^{\circ }$$

Check: $95^{\circ }+70^{\circ }+85^{\circ }+110^{\circ }=360^{\circ }$.

An everyday example. A circular garden with a straight path extended beyond one corner of a four-sided flower bed shows the exterior angle matching the far corner.

The link. This property is simply the supplementary-angle result combined with angles on a straight line.

**Show that a pair of opposite angles adds to $180^{\circ }$, or that an exterior angle equals the opposite interior angle, or that one segment subtends equal angles at two points on the same side of it.

Test 1 — supplementary opposite angles.** A quadrilateral has $\angle A=70^{\circ }$, $\angle B=95^{\circ }$, $\angle C=110^{\circ }$. Then $\angle D=360^{\circ }-275^{\circ }=85^{\circ }$, and $\angle A+\angle C=180^{\circ }$, so it is cyclic.

Test 2 — equal angles on the same segment. Points $P$ and $Q$ lie on the same side of $AB$ with $\angle APB=\angle AQB=40^{\circ }$. **So $A$, $B$, $P$ and $Q$ are concyclic.

Worked proof — isosceles trapezium.** In $ABCD$, $AB\parallel DC$ and $AD=BC$.

- The base angles are equal: $\angle A=\angle B$
- Co-interior angles: $\angle A+\angle D=180^{\circ }$
- **So $\angle B+\angle D=180^{\circ }$**, and $ABCD$ is cyclic

An everyday example. The four corners of any rectangular photo frame lie on one circle, because each angle is $90^{\circ }$ and opposite angles add to $180^{\circ }$.

The boundary case. A rhombus that is not a square is never cyclic, since its unequal opposite angles cannot add to $180^{\circ }$ while being equal to each other.

Write down every angle fact you can get from the figure — parallel lines, equal chords, radii and the cyclic properties — and chain them until the required angle appears, giving a reason for each step.

Worked example 1 — parallel lines. $ABCD$ is cyclic, $AB\parallel DC$ and $\angle BAD=100^{\circ }$.

$$\angle ADC=180^{\circ }-100^{\circ }=80^{\circ }\text{(co-interior)}\angle ABC=180^{\circ }-80^{\circ }=100^{\circ }\angle BCD=80^{\circ }$$

Worked example 2 — central angle. $\angle AOC=130^{\circ }$ with $D$ on the major arc and $B$ on the minor arc.

$$\angle ADC=65^{\circ }\angle ABC=180^{\circ }-65^{\circ }=115^{\circ }$$

Worked example 3 — equal chords. In cyclic $ABCD$, $AB=BC$ and $\angle ACB=30^{\circ }$.

$$\angle BAC=30^{\circ }\text{(isosceles)}\angle ABC=180^{\circ }-60^{\circ }=120^{\circ }\angle ADC=180^{\circ }-120^{\circ }=60^{\circ }$$

Check by another route: $\angle ADB=\angle ACB=30^{\circ }$ and $\angle BDC=\angle BAC=30^{\circ }$ (same segment), so $\angle ADC=60^{\circ }$.

An everyday example. A circular window with a four-sided stained-glass panel has angles that can all be found from one or two measured angles.

The substance. Equal chords subtend equal angles, which is often the hidden link in a rider.