Ptolemy’s Theorem

by aoum, Apr 16, 2025, 1:06 AM

Ptolemy’s Theorem: A Classic Result in Euclidean Geometry

Ptolemy’s Theorem is a foundational result in cyclic geometry that describes a precise relationship between the sides and diagonals of a cyclic quadrilateral (a quadrilateral inscribed in a circle). Named after the ancient Greek mathematician Claudius Ptolemy, it lies at the heart of trigonometry and classical geometry.

[asy]
size(200);
pair A = dir(110);
pair B = dir(40);
pair C = dir(-20);
pair D = dir(210);
draw(circle((0,0),1));
draw(A--B--C--D--cycle);
draw(A--C, dashed+blue);
draw(B--D, dashed+blue);
label("$A$", A, NW);
label("$B$", B, NE);
label("$C$", C, SE);
label("$D$", D, SW);
label("$AC$", midpoint(A--C), NE);
label("$BD$", midpoint(B--D), SW);
label("$AB$", midpoint(A--B), N);
label("$BC$", midpoint(B--C), E);
label("$CD$", midpoint(C--D), S);
label("$DA$", midpoint(D--A), W);
[/asy]
Ptolemy's theorem is a relation among these lengths in a cyclic quadrilateral.

1. Statement of Ptolemy’s Theorem

Let $ABCD$ be a cyclic quadrilateral. Then the product of the two diagonals equals the sum of the products of the two pairs of opposite sides:

$$
AC \cdot BD = AB \cdot CD + AD \cdot BC.
$$
This relation is exact if and only if the quadrilateral is inscribed in a circle.

2. Applications of Ptolemy’s Theorem
  • Deriving trigonometric identities
  • Proving special cases of triangle relationships
  • Solving geometry olympiad problems
  • Constructing chord lengths in circle geometry

3. Special Cases

If $ABCD$ is a rectangle (so $AB = CD$ and $AD = BC$), then the theorem reduces to the Pythagorean Theorem:

$$
AC^2 = AB^2 + AD^2.
$$
This is because in a rectangle inscribed in a circle, the diagonals are equal and intersect at right angles.

4. Proof Using Similar Triangles

Let’s outline a synthetic geometric proof:

Construct point $E$ on $AC$ such that $\angle EBD = \angle ABC$. Then triangles $EBD$ and $ABC$ are similar (angle-angle). Therefore:

$$
\frac{EB}{AB} = \frac{BD}{BC} \Rightarrow EB = \frac{AB \cdot BD}{BC}.
$$
Similarly, one shows:

$$
EC = \frac{AD \cdot BC}{AB}.
$$
Now since $E$ lies on $AC$, and $AC = EB + EC$, adding gives:

$$
AC = \frac{AB \cdot BD}{BC} + \frac{AD \cdot BC}{AB}.
$$
Multiplying both sides by $BD$ and simplifying yields the result:

$$
AC \cdot BD = AB \cdot CD + AD \cdot BC.
$$
5. Ptolemy's Inequality

If the quadrilateral $ABCD$ is not cyclic, the relation becomes an inequality:

$$
AC \cdot BD \le AB \cdot CD + AD \cdot BC,
$$
with equality if and only if the quadrilateral is cyclic.

6. Coordinate Proof (Analytic Geometry)

Place the cyclic quadrilateral on the unit circle using complex numbers:

Let $A = e^{i\alpha}$, $B = e^{i\beta}$, $C = e^{i\gamma}$, and $D = e^{i\delta}$ with $0 < \alpha < \beta < \gamma < \delta < 2\pi$. Then the chord lengths correspond to:

$$
|AC| = |e^{i\alpha} - e^{i\gamma}|, \quad |BD| = |e^{i\beta} - e^{i\delta}|.
$$
Ptolemy’s Theorem follows from identities involving complex absolute values and angles.

7. Trigonometric Form of Ptolemy’s Theorem

If $ABCD$ is inscribed in a circle, and the arcs subtended by sides are known, then the theorem can also be written using the law of sines:

$$
\sin(\angle A) \cdot \sin(\angle C) = \sin(\angle B) \cdot \sin(\angle D).
$$
8. Connections and Historical Note

Ptolemy used this theorem extensively in his work Almagest to create a table of chords — the precursor to modern trigonometric functions. In that setting, he used it to find exact values for chords of angles such as $36^\circ$ and $72^\circ$.

9. References

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