Cyclic Quadrilateral Theorem

A proof that opposite angles of a cyclic quadrilateral sum to 180°, using the inscribed angle theorem and arc addition.

Keywords: cyclic quadrilateral, inscribed quadrilateral, opposite angles, supplementary, inscribed angle, arc

Prerequisites: circle-angles-intro, inscribed-angle-proof · Difficulty: intermediate

Cyclic Quadrilateral Theorem

A cyclic quadrilateral is a quadrilateral whose four vertices all lie on a single circle.

Theorem: The opposite angles of a cyclic quadrilateral are supplementary (sum to $180°$):

\angle A + \angle C = 180° \qquad \angle B + \angle D = 180°
    

Proof Strategy

Place $ABCD$ on a circle with center $O$.

• $\angle A$ is an inscribed angle subtending arc $BCD$.
• $\angle C$ is an inscribed angle subtending arc $DAB$.
• Arcs $BCD$ and $DAB$ together make up the full circle ($360°$).

By the inscribed angle theorem:

\angle A = \tfrac{1}{2} \cdot \text{arc } BCD, \quad
      \angle C = \tfrac{1}{2} \cdot \text{arc } DAB
    

Adding:

\angle A + \angle C = \tfrac{1}{2}(360°) = 180°
    

Converse

If a quadrilateral has opposite angles summing to $180°$, then it is cyclic — a circle passes through all four vertices.

The Proof

Step 1: A Quadrilateral on a Circle

A cyclic quadrilateral has all four vertices on a circle. We draw quadrilateral $ABCD$ inscribed in a circle with center $O$.

Step 2: Opposite Angles Subtend Complementary Arcs

The key insight: opposite angles subtend arcs that together make the full circle.

• $\angle A$ (at vertex $A$) is an inscribed angle subtending arc $BCD$ — the arc from $B$ to $D$ passing through $C$.
• $\angle C$ (at vertex $C$) is an inscribed angle subtending arc $DAB$ — the arc from $D$ to $B$ passing through $A$.

These two arcs together comprise the entire circle ($360°$).

Step 3: Opposite Angles Sum to $180°$

By the inscribed angle theorem:

• $\angle A = \frac{1}{2} \cdot \text{arc } BCD$
• $\angle C = \frac{1}{2} \cdot \text{arc } DAB$

Since arc $BCD$ + arc $DAB$ = $360°$:

\angle A + \angle C = \tfrac{1}{2} \cdot 360° = 180°
    

By the same reasoning, $\angle B + \angle D = 180°$.

Step 4: The Converse

Converse: If a quadrilateral has opposite angles summing to $180°$, then it is cyclic.

Suppose $\angle A + \angle C = 180°$. Construct the circle through $A$, $B$, $D$. If $C$ were not on this circle, the inscribed angle at $C'$ on the circle would differ from $\angle C$, contradicting $\angle A + \angle C = 180°$. So $C$ must lie on the circle.

Conclusion

In a cyclic quadrilateral $ABCD$, opposite angles are supplementary:

\angle A + \angle C = 180°, \qquad \angle B + \angle D = 180°
    

The converse also holds: if a quadrilateral's opposite angles sum to $180°$, then it is cyclic.