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AMC 12 focus

Cyclic Quadrilaterals & Ptolemy

A quadrilateral inscribed in a circle carries extra structure — supplementary opposite angles, and a clean relationship between its sides and diagonals.

Supplementary opposite angles

In a cyclic quadrilateral ABCDABCD (all four vertices on one circle), opposite angles always sum to 180°180°:

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

This is often the fastest way to prove a quadrilateral is cyclic in the first place: show a pair of opposite angles is supplementary, and the converse guarantees the four points lie on a common circle.

A related fact: angles that subtend the same arc from the same side are equal — so ∠ACB=∠ADB\angle ACB = \angle ADBwhenever both angles look out at chord ABAB from the same arc.

Ptolemy's theorem

For a cyclic quadrilateral ABCDABCD, the product of the diagonals equals the sum of the products of opposite sides:

AC⋅BD=AB⋅CD+BC⋅ADAC \cdot BD = AB \cdot CD + BC \cdot AD

This is a genuinely powerful relationship — it lets you solve for an unknown diagonal or side using only lengths, with no angle computation at all, as long as you know the quadrilateral is cyclic.

Worked example

Problem: Cyclic quadrilateral ABCDABCD has AB=3AB=3, BC=4BC=4,CD=5CD=5, DA=6DA=6, and diagonal BD=7BD=7. Find diagonal ACAC.

Solution: By Ptolemy's theorem, AC⋅BD=AB⋅CD+BC⋅DAAC \cdot BD = AB\cdot CD + BC \cdot DA:

AC⋅7=3⋅5+4⋅6=15+24=39  ⟹  AC=397AC \cdot 7 = 3\cdot5 + 4\cdot6 = 15+24=39 \;\Longrightarrow\; AC = \dfrac{39}{7}

Practice problems

1.A cyclic quadrilateral has one angle of 72°72°. Find the angle opposite it.

2.A cyclic quadrilateral ABCDABCD has AB=CD=5AB=CD=5, BC=AD=6BC=AD=6, and diagonals AC=BD=xAC=BD=x. Use Ptolemy's theorem to find xx.

3.A quadrilateral has two pairs of opposite angles that are each supplementary. What can you conclude about it?

4.Square ABCDABCD has side length 4. Verify Ptolemy's theorem using its diagonals.

Angle ChasingTangency, Incircles & Excircles