Ptolemy's theorem says that for a cyclic quadrilateral, the product of opposing sides, summed, is equal to the product of the diagonals:

AB CD + BC AD = AC BD

Proof. adapted from wikipedia (here).

Let the angle s (red dot) subtend arc AB and the angle t (black dot) subtend arc CD. Then the central angle DPC = s + t and it has sin s + t. The other central angle APD has the same sine, as it is supplementary to s + t.

Let the components of the diagonals be AC = q + s and BD = p + r.

Twice the areas of the four small triangles will then be equal to

(pq + qr + rs + sp) sin s + t

Simple algebra will show that

(pq + qr + rs + sp) = (p + r)(q + s) = AC BD

The product of the diagonals times the sine of either central angle is equal to twice the area of the quadrilateral. We're on to something.

Now, the great idea. Move D to D', so that AD' = CD and CD' = AD.

The triangles ACD and ACD' are congruent, by SSS, so they have the same area.

Therefore the area of ABCD is equal to the area of ABCD'.

Some of the angles switch with the arcs. In particular, angle t (black dot) now subtends arc AD'. As a result s + t is the measure of the whole angle at vertex C. The whole angle at vertex A is supplementary, and the sine of the whole angle at vertex A is equal to that at C.

So twice the area of triangle ABD' is AB AD' sin s + t, and twice that of BCD' is BC CD' sin s + t. Add these two areas, equate them with the previous result, and factor out the common term sin s + t:

AC BD = AB AD' + BC CD'

But AD' = CD and CD' = AD so

AC BD = AB CD + BC AD

This is Ptolemy's theorem.

Here is one result from the theorem. Draw an equilateral triangle and its circumcircle. Pick any other point P on the circle and connect it to the vertices as shown.

Ptolemy says that

ps = qs + rs

p = q + r