GeometryDifficulty 35-8557 tagged problems

Ptolemy's Theorem

Ptolemy's Theorem is the habit of using cyclic quadrilateral side and diagonal products to unlock lengths and ratios. In contest math, that habit turns a crowded setup into a relation the student can test, bound, count, or compute. MathGrit teaches it as a recognizable signal, a deliberate move, and a final translation back to the original question.

Practice this technique Match my level

The Key Result

$$AC \cdot BD = AB \cdot CD + AD \cdot BC$$

For a cyclic quadrilateral, the product of the diagonals equals the sum of the products of the two pairs of opposite sides.

Why It Works

1
Let $ABCD$ be cyclic (in order). Choose point $P$ on diagonal $BD$ with $\angle BAP = \angle CAD$.
2
Then $\triangle ABP \sim \triangle ACD$ (equal angles from the construction and equal inscribed angles $\angle ABP = \angle ACD$), giving $AB\cdot CD = AC\cdot BP$.
3
Similarly $\triangle APD \sim \triangle ABC$, giving $AD\cdot BC = AC\cdot PD$.
4
Adding: $AB\cdot CD + AD\cdot BC = AC\,(BP + PD) = AC\cdot BD$, since $P$ lies on $BD$.

Worked Examples

Example 1

A point $P$ lies on the circumcircle of equilateral triangle $ABC$, on arc $BC$ not containing $A$. Prove $PA = PB + PC$.

1
$ABPC$ is a cyclic quadrilateral (with $P$ on arc $BC$), so apply Ptolemy: $PA \cdot BC = PB \cdot AC + PC \cdot AB$.
2
Since $\triangle ABC$ is equilateral, $AB = BC = CA = s$.
3
Substitute: $PA \cdot s = PB \cdot s + PC \cdot s$, and divide by $s$.
4
This gives $PA = PB + PC$.

Answer:

$PA = PB + PC$ (by Ptolemy on cyclic quadrilateral $ABPC$).

Warm-up

Apply Ptolemy's theorem to a rectangle $ABCD$ with $AB = CD = 5$ and $BC = AD = 12$ to find the diagonal.

1
A rectangle is cyclic (all four right angles, opposite angles sum to $180^\circ$), and its diagonals are equal: $AC = BD = d$.
2
Ptolemy: $AC\cdot BD = AB\cdot CD + AD\cdot BC$, i.e. $d^2 = 5\cdot 5 + 12\cdot 12 = 25 + 144 = 169$.
3
So $d = 13$ — exactly the Pythagorean theorem, recovered as a special case of Ptolemy.

Answer:

$d = 13$ (Ptolemy reduces to Pythagoras for a rectangle).

Contest level

An isosceles trapezoid $ABCD$ has parallel sides $AB = 6$ and $CD = 10$, with legs $BC = DA = 5$. Find the length of a diagonal.

1
An isosceles trapezoid is cyclic, and its two diagonals are equal: $AC = BD = e$.
2
Ptolemy: $AC\cdot BD = AB\cdot CD + AD\cdot BC$, so $e^2 = 6\cdot 10 + 5\cdot 5 = 60 + 25 = 85$.
3
Therefore $e = \sqrt{85}$. (Check by coordinates: $D=(0,0)$, $C=(10,0)$, $A=(2,\sqrt{21})$, $B=(8,\sqrt{21})$ give $AC = \sqrt{64 + 21} = \sqrt{85}$.)

Answer:

Each diagonal has length $\sqrt{85}$.

Olympiad / Challenge

In a regular pentagon $ABCDE$ with side length $1$, let $d$ be the length of a diagonal. Use Ptolemy on cyclic quadrilateral $ABCD$ to show $d = \dfrac{1 + \sqrt5}{2}$ (the golden ratio).

1
The pentagon is inscribed in a circle, so $ABCD$ is a cyclic quadrilateral. Its sides are $AB = BC = CD = 1$ (pentagon sides) and $DA$ is a diagonal of the pentagon, so $DA = d$.
2
Its diagonals are $AC$ and $BD$, each a diagonal of the pentagon, so $AC = BD = d$.
3
Ptolemy: $AC\cdot BD = AB\cdot CD + BC\cdot DA$, i.e. $d\cdot d = 1\cdot 1 + 1\cdot d$, giving $d^2 = d + 1$.
4
Solve $d^2 - d - 1 = 0$: $d = \dfrac{1 + \sqrt5}{2}$ (taking the positive root) $= \varphi$, the golden ratio.

Answer:

$d = \dfrac{1 + \sqrt5}{2} = \varphi$ — the diagonal-to-side ratio of a regular pentagon is the golden ratio.

Going Deeper

Generalization: dropping the cyclic hypothesis gives Ptolemy's INEQUALITY, $AC\cdot BD \le AB\cdot CD + AD\cdot BC$ for ANY four points, with equality exactly when $ABCD$ is a convex cyclic quadrilateral — so Ptolemy is simultaneously a metric identity and a concyclicity test. There is also a 'second' Ptolemy giving the ratio of the diagonals, $\dfrac{AC}{BD} = \dfrac{AB\cdot AD + CB\cdot CD}{BA\cdot BC + DA\cdot DC}$.
Where it appears: Ptolemy converts a circle plus four marked points into a clean length equation — ideal for AIME problems on cyclic quadrilaterals and for deriving the golden ratio, the regular-polygon diagonal lengths, and the addition formulas $\sin(x+y), \cos(x+y)$ (apply Ptolemy to a quadrilateral inscribed in a unit-diameter circle).
Pitfall: the vertices must be in CYCLIC ORDER $A, B, C, D$ around the circle, so that $AC$ and $BD$ are the genuine DIAGONALS and $AB, BC, CD, DA$ the sides. Pairing the wrong segments as 'diagonals' (e.g. taking a side as a diagonal because the labeling is scrambled) gives a false equation — always confirm the order around the circle first.

Spot the Signal

Look for problems where the key step is using cyclic quadrilateral side and diagonal products to unlock lengths and ratios.
You can describe the hard part as using cyclic quadrilateral side and diagonal products to unlock lengths and ratios, but a direct attack starts producing clutter.
The problem rewards preserving structure instead of expanding, listing, or guessing too early.

Learn the Move

Start by identify the diagram relation that calls for ptolemy's theorem, then rewrite the givens around it.
Name the relation that makes Ptolemy's Theorem legal before doing computation.
Use the new relation to replace the messiest part of the problem with a cleaner one.
Translate the result back to the quantity the problem actually asks for.

Avoid These Traps

Do not use Ptolemy's Theorem just because the surface looks familiar; verify the required condition first.
Applying Ptolemy's Theorem because it sounds relevant, without checking the trigger first.
Stopping after spotting the technique instead of finishing the calculation or proof.

MathGrit Coach Note

Let ptolemy's theorem reveal the geometric configuration; then compute only what remains.

Try it on:

Practice a contest problem where the key step is using cyclic quadrilateral side and diagonal products to unlock lengths and ratios.

Back to techniques

GeometryDifficulty 35-8557 tagged problems

Ptolemy's Theorem

Practice this technique Match my level

The Key Result

$$AC \cdot BD = AB \cdot CD + AD \cdot BC$$

For a cyclic quadrilateral, the product of the diagonals equals the sum of the products of the two pairs of opposite sides.

Why It Works

1
Let $ABCD$ be cyclic (in order). Choose point $P$ on diagonal $BD$ with $\angle BAP = \angle CAD$.
2
Then $\triangle ABP \sim \triangle ACD$ (equal angles from the construction and equal inscribed angles $\angle ABP = \angle ACD$), giving $AB\cdot CD = AC\cdot BP$.
3
Similarly $\triangle APD \sim \triangle ABC$, giving $AD\cdot BC = AC\cdot PD$.
4
Adding: $AB\cdot CD + AD\cdot BC = AC\,(BP + PD) = AC\cdot BD$, since $P$ lies on $BD$.

Worked Examples

Example 1

A point $P$ lies on the circumcircle of equilateral triangle $ABC$, on arc $BC$ not containing $A$. Prove $PA = PB + PC$.

1
$ABPC$ is a cyclic quadrilateral (with $P$ on arc $BC$), so apply Ptolemy: $PA \cdot BC = PB \cdot AC + PC \cdot AB$.
2
Since $\triangle ABC$ is equilateral, $AB = BC = CA = s$.
3
Substitute: $PA \cdot s = PB \cdot s + PC \cdot s$, and divide by $s$.
4
This gives $PA = PB + PC$.

Answer:

$PA = PB + PC$ (by Ptolemy on cyclic quadrilateral $ABPC$).

Warm-up

Apply Ptolemy's theorem to a rectangle $ABCD$ with $AB = CD = 5$ and $BC = AD = 12$ to find the diagonal.

1
A rectangle is cyclic (all four right angles, opposite angles sum to $180^\circ$), and its diagonals are equal: $AC = BD = d$.
2
Ptolemy: $AC\cdot BD = AB\cdot CD + AD\cdot BC$, i.e. $d^2 = 5\cdot 5 + 12\cdot 12 = 25 + 144 = 169$.
3
So $d = 13$ — exactly the Pythagorean theorem, recovered as a special case of Ptolemy.

Answer:

$d = 13$ (Ptolemy reduces to Pythagoras for a rectangle).

Contest level

An isosceles trapezoid $ABCD$ has parallel sides $AB = 6$ and $CD = 10$, with legs $BC = DA = 5$. Find the length of a diagonal.

1
An isosceles trapezoid is cyclic, and its two diagonals are equal: $AC = BD = e$.
2
Ptolemy: $AC\cdot BD = AB\cdot CD + AD\cdot BC$, so $e^2 = 6\cdot 10 + 5\cdot 5 = 60 + 25 = 85$.
3
Therefore $e = \sqrt{85}$. (Check by coordinates: $D=(0,0)$, $C=(10,0)$, $A=(2,\sqrt{21})$, $B=(8,\sqrt{21})$ give $AC = \sqrt{64 + 21} = \sqrt{85}$.)

Answer:

Each diagonal has length $\sqrt{85}$.

Olympiad / Challenge

In a regular pentagon $ABCDE$ with side length $1$, let $d$ be the length of a diagonal. Use Ptolemy on cyclic quadrilateral $ABCD$ to show $d = \dfrac{1 + \sqrt5}{2}$ (the golden ratio).

1
The pentagon is inscribed in a circle, so $ABCD$ is a cyclic quadrilateral. Its sides are $AB = BC = CD = 1$ (pentagon sides) and $DA$ is a diagonal of the pentagon, so $DA = d$.
2
Its diagonals are $AC$ and $BD$, each a diagonal of the pentagon, so $AC = BD = d$.
3
Ptolemy: $AC\cdot BD = AB\cdot CD + BC\cdot DA$, i.e. $d\cdot d = 1\cdot 1 + 1\cdot d$, giving $d^2 = d + 1$.
4
Solve $d^2 - d - 1 = 0$: $d = \dfrac{1 + \sqrt5}{2}$ (taking the positive root) $= \varphi$, the golden ratio.

Answer:

$d = \dfrac{1 + \sqrt5}{2} = \varphi$ — the diagonal-to-side ratio of a regular pentagon is the golden ratio.

Going Deeper

Generalization: dropping the cyclic hypothesis gives Ptolemy's INEQUALITY, $AC\cdot BD \le AB\cdot CD + AD\cdot BC$ for ANY four points, with equality exactly when $ABCD$ is a convex cyclic quadrilateral — so Ptolemy is simultaneously a metric identity and a concyclicity test. There is also a 'second' Ptolemy giving the ratio of the diagonals, $\dfrac{AC}{BD} = \dfrac{AB\cdot AD + CB\cdot CD}{BA\cdot BC + DA\cdot DC}$.
Where it appears: Ptolemy converts a circle plus four marked points into a clean length equation — ideal for AIME problems on cyclic quadrilaterals and for deriving the golden ratio, the regular-polygon diagonal lengths, and the addition formulas $\sin(x+y), \cos(x+y)$ (apply Ptolemy to a quadrilateral inscribed in a unit-diameter circle).
Pitfall: the vertices must be in CYCLIC ORDER $A, B, C, D$ around the circle, so that $AC$ and $BD$ are the genuine DIAGONALS and $AB, BC, CD, DA$ the sides. Pairing the wrong segments as 'diagonals' (e.g. taking a side as a diagonal because the labeling is scrambled) gives a false equation — always confirm the order around the circle first.

Spot the Signal

Look for problems where the key step is using cyclic quadrilateral side and diagonal products to unlock lengths and ratios.
You can describe the hard part as using cyclic quadrilateral side and diagonal products to unlock lengths and ratios, but a direct attack starts producing clutter.
The problem rewards preserving structure instead of expanding, listing, or guessing too early.

Learn the Move

Start by identify the diagram relation that calls for ptolemy's theorem, then rewrite the givens around it.
Name the relation that makes Ptolemy's Theorem legal before doing computation.
Use the new relation to replace the messiest part of the problem with a cleaner one.
Translate the result back to the quantity the problem actually asks for.

Avoid These Traps

Do not use Ptolemy's Theorem just because the surface looks familiar; verify the required condition first.
Applying Ptolemy's Theorem because it sounds relevant, without checking the trigger first.
Stopping after spotting the technique instead of finishing the calculation or proof.

MathGrit Coach Note

Let ptolemy's theorem reveal the geometric configuration; then compute only what remains.

Try it on:

Practice a contest problem where the key step is using cyclic quadrilateral side and diagonal products to unlock lengths and ratios.