Editing Ptolemy's theorem (section)

=== 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.