Editing Ptolemy's theorem (section)

==Proofs==
{{Further |Proofs of trigonometric identities}}

=== Visual proof ===

[[File:Animated visual proof of Ptolemy's theorem, based on Derrick & Herstein (2012).gif|thumb|Animated visual proof of Ptolemy's theorem, based on Derrick & Herstein (2012).]]
The animation here shows a visual demonstration of Ptolemy's theorem, based on Derrick & Herstein (2012).<ref>W. Derrick, J. Herstein (2012) Proof Without Words: Ptolemy's Theorem, The College Mathematics Journal, v.43, n.5, p. 386.</ref>

=== Proof by similarity of triangles ===
[[Image:Ptolemy's theorem.svg|thumb|Constructions for a proof of Ptolemy's theorem]]
Let ABCD be a [[cyclic quadrilateral]].
On the [[chord (geometry)|chord]] BC, the [[inscribed angle]]s ∠BAC = ∠BDC, and on AB, ∠ADB = ∠ACB.
Construct K on AC such that ∠ABK = ∠CBD; since ∠ABK + ∠CBK = ∠ABC = ∠CBD + ∠ABD, ∠CBK = ∠ABD.

Now, by common angles △ABK is [[Similarity (geometry)|similar]] to △DBC, and likewise △ABD is similar to △KBC.
Thus AK/AB = CD/BD, and CK/BC = DA/BD;
equivalently, AK⋅BD = AB⋅CD, and CK⋅BD = BC⋅DA.
By adding two equalities we have AK⋅BD + CK⋅BD = AB⋅CD + BC⋅DA, and factorizing this gives (AK+CK)·BD = AB⋅CD + BC⋅DA.
But AK+CK = AC, so AC⋅BD = AB⋅CD + BC⋅DA, [[Q.E.D.]]<ref>{{citation|title=Charming Proofs: A Journey Into Elegant Mathematics|volume=42|series=Dolciani Mathematical Expositions|first1=Claudi|last1=Alsina|first2=Roger B.|last2=Nelsen|publisher=[[Mathematical Association of America]]|year=2010|isbn=9780883853481|page=112|url=https://books.google.com/books?id=mIT5-BN_L0oC&pg=PA112}}.</ref>

The proof as written is only valid for [[simple polygon|simple]] cyclic quadrilaterals. If the quadrilateral is self-crossing then K will be located outside the line segment AC. But in this case, AK&minus;CK = ±AC, giving the expected result.

=== Proof by trigonometric identities ===
Let the inscribed angles subtended by <math>AB</math>, <math>BC</math> and <math>CD</math> be, respectively, <math>\alpha</math>, <math>\beta</math> and <math>\gamma</math>, and the radius of the circle be <math>R</math>, then we have <math>AB=2R\sin\alpha</math>, <math>BC=2R\sin\beta</math>, <math>CD=2R\sin\gamma</math>, <math>AD=2R\sin(180^\circ-(\alpha+\beta+\gamma))</math>, <math>AC=2R\sin(\alpha+\beta)</math> and <math>BD=2R\sin(\beta+\gamma)</math>, and the original equality to be proved is transformed to

:<math> \sin(\alpha+\beta)\sin(\beta+\gamma) = \sin\alpha\sin\gamma + \sin\beta \sin(\alpha + \beta+\gamma)</math>

from which the factor <math>4R^2</math> has disappeared by dividing both sides of the equation by it.

Now by using the sum formulae, <math>\sin(x+y)=\sin{x}\cos y+\cos x\sin y</math> and <math>\cos(x+y)=\cos x\cos y-\sin x\sin y</math>, it is trivial to show that both sides of the above equation are equal to

:<math>
\begin{align}
& \sin\alpha\sin\beta\cos\beta\cos\gamma + \sin\alpha\cos^2\beta\sin\gamma \\ + {} & \cos\alpha\sin^2\beta\cos\gamma+\cos\alpha\sin\beta\cos\beta\sin\gamma.
\end{align}
</math>

Q.E.D.

Here is another, perhaps more transparent, proof using rudimentary trigonometry.
Define a new quadrilateral <math>ABCD'</math> inscribed in the same circle, where <math>A,B,C</math> are the same
as in <math>ABCD</math>, and <math>D'</math> located at a new point on the same circle, defined by <math> |\overline{AD'}| = |\overline{CD}|</math>, 
<math>|\overline{CD'}| = |\overline{AD}|</math>.  (Picture triangle <math>ACD</math> flipped, so that vertex <math>C</math> moves to vertex <math>A</math> and vertex <math>A</math> moves to vertex <math>C</math>.  Vertex <math>D</math> will now be located at a new point D’ on the circle.)
Then, <math>ABCD'</math> has the same edges lengths, and consequently the same inscribed angles subtended by 
the corresponding edges, as <math> ABCD</math>, only in a different order. That is, <math>\alpha</math>, <math>\beta</math> and <math>\gamma</math>, for,  respectively, <math>AB, BC</math> and <math>AD'</math>.
Also, <math>ABCD</math> and <math>ABCD'</math> have the same area. Then,

:<math> 
\begin{align}
\mathrm{Area}(ABCD) & = \frac{1}{2}  AC\cdot BD \cdot \sin(\alpha + \gamma); \\
\mathrm{Area}(ABCD') & = \frac{1}{2} AB\cdot AD'\cdot \sin(180^\circ - \alpha - \gamma) +
 \frac{1}{2} BC\cdot CD' \cdot \sin(\alpha + \gamma)\\ & = 
\frac{1}{2} (AB\cdot CD +  BC\cdot AD)\cdot \sin(\alpha + \gamma).
\end{align}
</math>
 
Q.E.D.

=== Proof by inversion ===
[[File:Ptolemy-crop.svg|thumb|Proof of Ptolemy's theorem via [[circle inversion]]]]

Choose an auxiliary circle <math> \Gamma </math> of radius <math> r </math> centered at D with respect to which the circumcircle of ABCD is [[Inversive geometry#Circle inversion|inverted]] into a line (see figure).
Then
<math> A'B' + B'C' = A'C'. </math>
Then <math> A'B', B'C' </math> and <math> A'C' </math> can be expressed as 
<math display="inline"> \frac{AB \cdot DB'}{DA} </math>, <math display="inline"> \frac{BC \cdot DB'}{DC} </math> and <math display="inline"> \frac{AC \cdot DC'}{DA} </math> respectively. Multiplying each term by <math display="inline"> \frac{DA \cdot DC}{DB'} </math> and using <math display="inline">\frac{DC'}{DB'} = \frac{DB}{DC} </math> yields Ptolemy's equality.

Q.E.D.

Note that if the quadrilateral is not cyclic then A', B' and C' form a triangle and hence A'B'+B'C' > A'C', giving us a very simple proof of Ptolemy's Inequality which is presented below.

=== Proof using complex numbers ===
Embed ABCD in the [[complex plane]] <math>\mathbb{C}</math> by identifying <math>A\mapsto z_A,\ldots,D\mapsto z_D</math> as four distinct [[complex number]]s <math>z_A,\ldots,z_D\in\mathbb{C}</math>. Define the [[cross-ratio]]
:<math>\zeta:=\frac{(z_A-z_B)(z_C-z_D)}{(z_A-z_D)(z_B-z_C)}\in\mathbb{C}_{\neq0}</math>.
Then 
:<math>
\begin{align}
\overline{AB}\cdot\overline{CD}+\overline{AD}\cdot\overline{BC}
& = \left|z_A-z_B\right|\left|z_C-z_D\right| + \left|z_A-z_D\right|\left|z_B-z_C\right| \\
& = \left|(z_A-z_B)(z_C-z_D)\right| + \left|(z_A-z_D)(z_B-z_C)\right| \\
& = \left(\left|\frac{(z_A-z_B)(z_C-z_D)}{(z_A-z_D)(z_B-z_C)}\right| + 1\right) \left|(z_A-z_D)(z_B-z_C)\right| \\
& = \left(\left|\zeta\right| +1\right) \left|(z_A-z_D)(z_B-z_C)\right| \\
& \geq  \left|(\zeta +1)(z_A-z_D)(z_B-z_C)\right| \\
& = \left|(z_A-z_B)(z_C-z_D)+(z_A-z_D)(z_B-z_C)\right| \\
& = \left|(z_A-z_C)(z_B-z_D)\right| \\
& = \left|z_A-z_C\right|\left|z_B-z_D\right| \\
& = \overline{AC}\cdot\overline{BD}
\end{align}
</math>
with equality if and only if the cross-ratio <math>\zeta</math> is a positive real number.  This proves [[Ptolemy's inequality]] generally, as it remains only to show that <math>z_A,\ldots,z_D</math> lie consecutively arranged
on a circle (possibly of infinite radius, i.e. a line) in <math>\mathbb{C}</math> if and only if <math>\zeta\in\mathbb{R}_{>0}</math>.

From the [[Complex number#Polar form|polar form]] of a complex number <math>z=\vert z\vert e^{i\arg(z)}</math>, it follows
:<math>
\begin{align}
\arg(\zeta) & = \arg\frac{(z_A-z_B)(z_C-z_D)}{(z_A-z_D)(z_B-z_C)} \\
& = \arg(z_A-z_B)+\arg(z_C-z_D)-\arg(z_A-z_D)-\arg(z_B-z_C) \pmod{2\pi} \\
& = \arg(z_A-z_B)+\arg(z_C-z_D)-\arg(z_A-z_D)-\arg(z_C-z_B) - \arg(-1) \pmod{2\pi} \\
& = - \left[\arg(z_C-z_B)-\arg(z_A-z_B)\right] - \left[\arg(z_A-z_D)-\arg(z_C-z_D)\right] -\arg(-1) \pmod{2\pi} \\
& = - \angle ABC - \angle CDA -\pi \pmod{2\pi}\\
& = 0
\end{align}
</math>
with the last equality holding if and only if ABCD is cyclic, since a quadrilateral is cyclic if and only if opposite angles sum to <math>\pi</math>.

Q.E.D.

Note that this proof is equivalently made by observing that the cyclicity of ABCD, i.e. the [[Angle#Supplementary angle|supplementarity]] <math>\angle ABC</math> and <math>\angle CDA</math>, is equivalent to the condition
:<math>\arg\left[(z_A-z_B)(z_C-z_D)\right] = \arg\left[(z_A-z_D)(z_B-z_C)\right] = \arg\left[(z_A-z_C)(z_B-z_D)\right]  \pmod{2\pi}</math>;
in particular there is a rotation of <math>\mathbb{C}</math> in which this <math>\arg</math> is 0 (i.e. all three products are positive real numbers), and by which Ptolemy's theorem 
:<math>\overline{AB}\cdot \overline{CD}+\overline{AD}\cdot\overline{BC} = \overline{AC}\cdot \overline{BD}</math>
is then directly established from the simple algebraic identity
:<math>(z_A-z_B)(z_C-z_D)+(z_A-z_D)(z_B-z_C)=(z_A-z_C)(z_B-z_D).</math>