📋 Prerequisites

• Circles Basics
• Congruence & Similarity

A quadrilateral whose four vertices all lie on a circle is cyclic. These quadrilaterals have beautiful properties that competition problems love — opposite angles sum to 180°, the exterior angle equals the interior opposite, and Ptolemy's theorem relates all four sides and the diagonals.

📚 Key Concepts

What Is a Cyclic Quadrilateral?

A quadrilateral ABCD is cyclic (or concyclic) if all four vertices lie on a single circle (the circumcircle).

All triangles are cyclic (every triangle has a circumcircle), but not all quadrilaterals are.

Core Properties of Cyclic Quadrilaterals

1. Opposite angles sum to 180° (supplementary):
   $ \angle A + \angle C = 180° $ and $ \angle B + \angle D = 180° $
   Why: each pair intercepts arcs that together make the full circle (360°), and inscribed angles are half the arc.
2. Exterior angle = interior opposite angle:
   If you extend one side, the exterior angle equals the interior angle at the opposite vertex.
3. Angles subtended by the same chord are equal:
   $ \angle BAC = \angle BDC $ (both subtend arc BC)

How to Prove Four Points Are Concyclic

Any one of these conditions is enough:

1. Opposite angles supplementary: If $ \angle A + \angle C = 180° $, then ABCD is cyclic
2. Exterior angle = interior opposite: Extend a side, if the exterior angle equals the opposite interior angle
3. Equal angles subtended by the same chord: If points C and D are on the same side of AB and $ \angle ACB = \angle ADB $
4. Power of a point (converse): If two chords intersect and satisfy the power-of-point relationship
5. Ptolemy's inequality becomes equality: If $ AB \cdot CD + BC \cdot DA = AC \cdot BD $

Ptolemy's Theorem

For a cyclic quadrilateral with sides $a, b, c, d$ and diagonals $p, q$:

\[ ac + bd = pq \]

More precisely: $ AB \cdot CD + BC \cdot DA = AC \cdot BD $

The product of the diagonals equals the sum of the products of opposite sides.

Ptolemy's inequality: For any quadrilateral (not necessarily cyclic), $ ac + bd \geq pq $, with equality iff cyclic.

Brahmagupta's Formula

Area of a cyclic quadrilateral with sides $a, b, c, d$:

\[ \text{Area} = \sqrt{(s-a)(s-b)(s-c)(s-d)} \]

Where $ s = \frac{a+b+c+d}{2} $ (the semiperimeter).

This is the generalization of Heron's formula — set $d = 0$ and you get Heron's formula for a triangle!

Special Case: Cyclic Trapezoid

If a trapezoid is cyclic, it must be isosceles (the non-parallel sides are equal).

Conversely: every isosceles trapezoid is cyclic.

Why: parallel chords cut off equal arcs between them, making the legs equal.

Special Case: Right Kite

A kite with two opposite right angles is cyclic (the diagonal connecting the right angles is the diameter by Thales' theorem).

✏️ Example Problems