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Ptolemy's Theorem

How to use
  1. ▶ Play runs the proof, Next → and ← Back step one scene at a time, Reset restores the opening scene and the original figure. Learn more about the controls
  2. Drag BB, CC or DD round the circle, AA for the radius or OO for the centre; the readout under the canvas shows AC⋅BDAC \cdot BD and AB⋅CD+AD⋅BCAB \cdot CD + AD \cdot BC. Learn more about why it holds
  3. Labels toggles the point names, Measures writes each marked angle's size beside its arc. The Step explanations log keeps every step so far. Learn more about reading a scene
Proof?Take K on the diagonal AC with ∠ABK = ∠DBC. The inscribed angles on the arc BC make ∠BAC = ∠BDC, so ABK ~ DBC and AK·BD = AB·CD; the inscribed angles on the arc AB make ∠BCA = ∠BDA, so KBC ~ ABD and KC·BD = AD·BC. Add: AC·BD = AB·CD + AD·BC. Slide a vertex round the circle.Ptolemy's theorem
Display
Drag B, C or D round the circle, A for the radius, O for the centre
OABCD
AC·BD = 24.91, AB·CD + AD·BC = 24.91
Step 1 / 6

Step explanations

1A cyclic quadrilateral and its diagonals
Given
ABCD is inscribed in a circle, with diagonals AC and BD. To prove: AC · BD = AB · CD + AD · BC, the product of the diagonals equals the sum of the products of the two pairs of opposite sides.
Drag a vertex round the circle. Learn more about the statement

Getting Started with the Visualizer
The Statement
The Point K
Inscribed Angles
The First Pair of Similar Triangles
The Second Pair of Similar Triangles
Why It Holds
Dragging the Figure
Consequences



Getting Started with the Visualizer

DemoA run through the proof player
Step 0 of 7
Step through the proof, then drag BB, CC or DD round the circle and watch the two sides of the equation in the readout change together and stay equal. The picture below is the player with its parts numbered; the other tabs list what can be dragged, what each control does and what the step log shows.
Opening scene: the controls
1234567
1Proof — the theorem loaded in the player; the ? gives its idea in one line
2Display — Labels shows the point names, Measures writes each marked angle's size beside its arc
3Draggable points — B, C and D slide round the circle, A sets the radius, O moves the circle; the hollow point K follows on AC
4Readout — the line under the figure: AC·BD and AB·CD + AD·BC, in units of 60 pixels, live while you drag
5Back, Play, Next, Reset — step through the proof or run it; Reset returns to step 1 and restores the original figure
6Speed and step counter — how fast Play advances, and which of the 6 steps is on screen
7Step explanations — every step reached so far, with its justification tag and a link to the section that explains it; click an entry to jump back to it
Proof complete: step 6 of 6
12345
1The point K — on the diagonal AC; the two shaded pairs meet along BK
2The two pairs — ABK with DBC, and KBC with ABD, four shaded triangles over the quadrilateral
3The marked angles — θ twice at B, φ at A and D, ψ at C and D
4Justification tags — Given, Construction, Inscribed angles, AA similarity (twice), Adding
5Replay — at the last step Play becomes Replay and Next is off

The Statement

Ptolemy's theorem is about a quadrilateral ABCDABCD inscribed in a circle, a cyclic quadrilateral. It says that the product of the two diagonals equals the sum of the products of the two pairs of opposite sides:

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


It comes from Ptolemy's Almagest, where it was the tool for computing tables of chords, the ancestor of the sine table. The converse holds as well: a quadrilateral whose sides and diagonals satisfy this equation is cyclic; for every other quadrilateral the left side is smaller than the right (Ptolemy's inequality).

The proof uses the inscribed angle theorem and AA similarity twice, after one construction: a point KK on the diagonal ACAC chosen to copy an angle. In the player BB, CC and DD slide round the circle, AA sets the radius and OO moves the circle; the readout shows the two sides of the equation, in units of 60 pixels.
OABCD
Opening scene, frozen

The cyclic quadrilateral ABCD with its diagonals AC and BD.


The Point K

Everything turns on one well-chosen point. On the diagonal ACAC take the point KK for which

∠ABK=∠DBC\angle ABK = \angle DBC


Both angles are marked θ\theta in the player. The point exists because the ray BDBD runs inside the angle ABCABC, so ∠DBC\angle DBC is smaller than ∠ABC\angle ABC and the ray from BB making the angle θ\theta with BABA falls inside the angle, crossing ACAC. Taking θ\theta away from ∠ABC\angle ABC on either side gives a second pair of equal angles for free:

∠KBC=∠ABC−θ=∠ABD\angle KBC = \angle ABC - \theta = \angle ABD
θθOABCDK
The point K, frozen

K on AC with ∠ABK = ∠DBC, both marked θ.


Inscribed Angles

Two pairs of equal angles come from the circle. The angles ∠BAC\angle BAC and ∠BDC\angle BDC both stand on the arc BCBC, seen from AA and from DD on the far side of it; by the inscribed angle theorem they are equal, marked φ\varphi. In the same way ∠BCA\angle BCA and ∠BDA\angle BDA both stand on the arc ABAB and are equal, marked ψ\psi:

∠BAC=∠BDC=φ,∠BCA=∠BDA=ψ\angle BAC = \angle BDC = \varphi, \qquad \angle BCA = \angle BDA = \psi


Note that ∠BAK\angle BAK is the same angle as ∠BAC\angle BAC, and ∠BCK\angle BCK the same as ∠BCA\angle BCA, since KK lies on ACAC.
θθφφψψOABCDK
Inscribed angles, frozen

φ at A and D on the arc BC; ψ at C and D on the arc AB.


The First Pair of Similar Triangles

Compare the triangles ABKABK and DBCDBC. They have ∠ABK=∠DBC=θ\angle ABK = \angle DBC = \theta by the choice of KK, and ∠BAK=∠BDC=φ\angle BAK = \angle BDC = \varphi by the inscribed angles. By AA similarity they are similar, with A↔DA \leftrightarrow D, B↔BB \leftrightarrow B, K↔CK \leftrightarrow C, so corresponding sides are in proportion:

AKDC=ABDB,AK⋅BD=AB⋅CD\frac{AK}{DC} = \frac{AB}{DB}, \qquad AK \cdot BD = AB \cdot CD


One of the two products of opposite sides has appeared, paired with the piece AKAK of the diagonal.
θθφφψψOABCDK
The first pair, frozen

ABK and DBC shaded: AK·BD = AB·CD.


The Second Pair of Similar Triangles

Now compare KBCKBC and ABDABD. They have ∠KBC=∠ABD\angle KBC = \angle ABD (both are ∠ABC−θ\angle ABC - \theta) and ∠BCK=∠BDA=ψ\angle BCK = \angle BDA = \psi. Again by AA they are similar, with K↔AK \leftrightarrow A, B↔BB \leftrightarrow B, C↔DC \leftrightarrow D:

KCAD=BCBD,KC⋅BD=AD⋅BC\frac{KC}{AD} = \frac{BC}{BD}, \qquad KC \cdot BD = AD \cdot BC


The other product of opposite sides, paired with the other piece KCKC of the diagonal.
θθφφψψOABCDK
The second pair, frozen

KBC and ABD shaded: KC·BD = AD·BC.


Why It Holds

Add the two results:

AK⋅BD+KC⋅BD=AB⋅CD+AD⋅BCAK \cdot BD + KC \cdot BD = AB \cdot CD + AD \cdot BC


On the left, (AK+KC)⋅BD(AK + KC) \cdot BD, and AK+KC=ACAK + KC = AC because KK lies on the diagonal ACAC. So

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


The end-of-proof mark ∎ in the log closes the argument. The tags name the facts used: the given figure, one construction, the inscribed angle theorem, AA similarity twice, and an addition. The whole weight of the proof is carried by the point KK: it cuts the diagonal ACAC into two pieces, each of which is matched with one product of opposite sides.

For a quadrilateral that is not cyclic the same construction still gives two similar triangles, but KK no longer lies on ACAC; the triangle inequality AK+KC>ACAK + KC > AC then turns the equation into Ptolemy's inequality.
θθφφψψOABCDK
The conclusion, frozen

Adding, with AK + KC = AC: AC·BD = AB·CD + AD·BC.


Dragging the Figure

Drag BB, CC or DD round the circle: the quadrilateral changes shape, the point KK slides along ACAC to keep ∠ABK=∠DBC\angle ABK = \angle DBC, the four marked angles change, and the two sides of the equation in the readout change together and stay equal. Drag AA to change the radius (the other three vertices keep their places on the circle) or OO to move the whole figure.

The player keeps the four vertices in the order AA, BB, CC, DD round the circle, so the quadrilateral stays convex with ACAC and BDBD as its diagonals, keeps the centre inside it and keeps every side long enough for its labels. A drag that would break one of these is refused.
θθφφψψOABCDK
After dragging B, frozen

B slid towards C: AB longer, BC shorter, the two sides of the equation still equal.


Consequences

• Pythagoras. A rectangle is cyclic, with equal diagonals dd and sides aa, bb: Ptolemy gives d2=a2+b2d^2 = a^2 + b^2, another proof of the Pythagorean theorem.
• The addition formulas. With ACAC a diameter of a circle of diameter 1, the sides are sines of the inscribed angles, and Ptolemy's equation becomes sin⁡(α+β)=sin⁡αcos⁡β+cos⁡αsin⁡β\sin(\alpha + \beta) = \sin\alpha\cos\beta + \cos\alpha\sin\beta; that is how Ptolemy built his table of chords.
• The regular pentagon. For four vertices of a regular pentagon with side ss and diagonal dd, Ptolemy gives d2=s2+sdd^2 = s^2 + sd, so d/sd/s is the golden ratio.
• Equilateral triangle. For a point PP on the arc BCBC of the circumcircle of an equilateral triangle ABCABC, Ptolemy gives PA=PB+PCPA = PB + PC.
• Ptolemy's inequality and the converse. For any four points AC⋅BD≤AB⋅CD+AD⋅BCAC \cdot BD \le AB \cdot CD + AD \cdot BC, with equality exactly when they lie on a circle (or a line) in that order; so the equation is a test for a quadrilateral to be cyclic, alongside the opposite angles test.
• Power of a point. The intersecting chords theorem is the other classical relation between the diagonals of a cyclic quadrilateral; together with Ptolemy it determines the two pieces of each diagonal from the four sides.