Every (non-degenerate) triangle has a
circumcircle – that is, there is a (unique) circle that passes through
all three corners. The same is not usualy true of a quadrilateral: pick any
three of the corners of your quardilateral and construct the circumcircle of the
triangle that has these as its corners; typically the fourth corner of your
quadrilateral won't be on that circle. However, when it does, the quadrilateral
does indeed have a circumcircle; such a quadrilateral is described
as cyclic

(a word that's used to mean other things in
other contexts, but geometers use it with this meaning for any polygon whose
vertices all lie on one circle).

Cyclic quadrilaterals have some neat properties, so let's take a look at a few results about them. For the first, the quadrilateral is implicit: it's the convex hull of the ends of the two chords, which are its diagonals.

Consider two chords that cross each other. Each cuts the other into two pieces;
what I'll be showing is that a rectangle whose sides are the two pieces of
either chord has the same area as the corresponding rectangle made of the pieces
of the other chord. (Representative rectangles are here shown
in pink and orange, along with quarter-turn arcs about the point of
intersection to show which side of each is equal to the other part of a chord;
the two rectangles have equal areas). This is known as The Intersecting
Chords Theorem

.

Let's have a second copy of the diagram, without the rectangles, to decorate with construction lines and markings, to aid the proof; we can start with labels on the parts of the intersecting chords, indicating their lengths, splitting one as p+q and the other as r+s; the theorem claims p.q = r.s. If we traverse the circle noting the order in which we meet the ends of our chords, we alternate chords, since the two chords cross each other. We can draw a bounding quadrilateral that joins up the four chord-ends, in the same order as the circle visits them; the intersecting chords are diagonals of this quadrilateral. I've drawn that (in grey) with one pair of opposite sides dashed and the other pair solid. Each side of this quadrilateral is a chord of the circle, of course; so angles subtended by it at the other two corners of the quadrilateral must be equal; I've labelled the angles opposite dashed sides with letters; equal angles have the same letter.

This shows that the triangle, formed by either solid side and the two parts of the intersecting chords on its side of the intersection, has the same angles in two of its corners as the other solid side's matching triangle; and the third angle in each triangle is just the angle (the same on both sides) between the intersecting chords where they meet. Thus these two triangles are (mirrored) similar and the lenghts of their corresponding edges are in a common proportion; the two solid sides of our bounding quadrilateral correspond, for these purposes; edges from the intersection of the original chords correspond precisely if their other ends meet that solid side in the same angle. Thus each triangle's part of one intersecting chord corresponds to the other triangle's part of the other chord; the ratios of the corresponding parts being equal, p/r = s/q, then implies that the product of the two parts into which either intersecting chord is cut is the same as the product for the two parts of the other chord, p.q = r.s, which is exactly what the theorem claims.

How about the converse ? Whenever two line segments cross each other and the products of lengths of the pieces into which they cut each other are equal, we can construct the circumcircle of the triangle formed by any three of the end-points. Since we've left out only one of the four end-points, both end-points of one of the crossing lines lie on the circle, as does one end-point of the other; the crossing-point of the two lines is in the interior of the line segment that has both ends on the circle; so construct a line from the third point, that's on the circle by construction, through the crossing-point until it hits the circle again; this makes a second chord crossing the first, to which we can apply the intersecting chords theorem, revealing that the extension past the crossing point was exactly by the distance from the crossing-point to the fourth end-point of the original line segments, which is thus the other end of the chord, so lies on the circle.

So, indeed, whenever two line segments cross, cutting each other into pieces whose products of lengths are equal, the circle through any three of the end-points must pass through the fourth as well.

When we apply the intersecting chords theorem to a chord and its perpendicular bisector chord, which is necessarily a diameter, the two halves of the bisected chord give us a square whose area is equal to that of a rectangle whose sides are the two parts into which it cuts the diameter. By constructing the diameter out of two sides of a rectangle, the intersecting chord theorem gives us a way to take the square root of a rectangle's area.

Now consider the special case where the two chords are perpendicular, still with one split into parts p, q by the other, which this first splits into parts r, s, with p.q = r.s. If we half-turn the p+q chord about the centre of the circle, it now splits the r+s chord as s+r and the distance between the original and rotated copy is r −s or s −r, whichever is positive. The diameter of the circle then has square (p+q).(p+q) +(r−s).(r−s) = p.p +q.q +r.r +s.s since 2.p.q −2.r.s = 0; so the areas of the squares on the four parts of our two chords add up to the same area as that of the square that bounds the circle; multiply this area by π/4 to get the area of the circle itself. For an illustration, see Catriona Shearer's puzzle that brought this to my attention.

Let's first look at what the angles opposite chords tell us: as shown in the diagram here, the two internal angles opposite each edge are equal.

We can use the sine formula for a chord's length to obtain the lengths of the sides and diagonals: each is just the diameter of the circle times the sine of the angle opposite it. This can actually be used to obtain Ptolemy's theorem, below, using the formulæ for the sines and cosines of sums of angles; but I prefer a more geometric proof. The combination of trigonometry and algebra is immensely powerful, but more geometric proofs tend to give more insight into what's going on.

To prepare for Ptolemy's theorem, though, let's look at a tangent to a circle
and the extensions, to where they cut it, of chords through the point opposite
where the tangent touches. We'll first see how the length of the chord and of
this extension are related; then how the lengths of chords *not* through
the opposite point are related to their projections onto the tangent by rays
through their end-points, from the point opposite the tangent.

First let's look at a chord from the point P opposite T, where the tangent touches the circle, through S on the circle and extend it to W on the tangent. The diameter PT meets the tangent in a right angle; and the angle PST is subtended by a diameter, so it's also a right angle. We thus get two right-angle triangles sharing the angle SPT, from which we can infer that PS/PT = PT/PW, hence PS.PW = PT.PT, the square on the diameter. The product of a chord, with one end opposite the tangency, and its extension to meet the tangent is the same for all chords: the square of the diameter (itself, indeed, both a chord and its own extension to meet the tangent).

Now let's look at a chord not through the point opposite to tangency, QR, and the line-segment along the tangent between where it's crossed by the rays from P through the two ends of our chord. From what we just showed, PQ.PU and PR.PV are both equal to the squared diameter; in particular, they're equal to each other, so PQ/PR = PV/PU. Each of these is a ratio of lengths along the same pair of lines out of P, albeit swapping the two lines between the ratios; but this means that PQR and PVU are similar triangles, albeit mirrored as well as scaled, hence that QR/UV = PQ/PV = PR/PU. Let D be the square on our diameter, to make PV = D/PR and PU = D/PQ, and we get QR/UV = PQ.PR/D from either version of this formula; or UV = QR.D/PQ/PR. The length along the tangent of a chord's projection onto it is the length of the chord times the squared diameter, divided by the product of distances of the chord's ends from the point opposite the tangency.

With these two results in hand, we're ready for a nice elegant proof of Ptolemy's theorem. It's not one I came up with for myself: it's adapted from one that Zvezdelina Stankova gave on Numberphile, that's motivated by central inversion; I just refactored it to skip the inversion. The results just obtained are all we needed from that. The same talk also introduced me to the equilateral triangle and pentagon results noted below

So, back to that rectangle we saw earlier, but now showing a tangent to its
circle opposite one of its vertices, and the various lines out of that vertex
extended to where they hit the tangent. We can necessarily do this, since the
whole circle lies on this
side

of the (parallel) tangent *at P*, so each of the rays here is
from P to a point definitely closer to the tangent we need it to hit; so, if
extended far enough, each ray surely shall eventually hit the tangent.

Along the tangent, we naturally have VW = VU +UW; and each of these three lengths is the projection onto the tangent of a chord not through P, so we can relate its length to that of the chord it's a projection of in terms of lengths along the rays out of P; and these can all be reduced to lengths of chords, as above, and the squared diagonal D. We get:

- VW = RS.D/PR/PS
- VU = QR.D/PQ/PR
- UW = QS.D/PQ/PS

Multiplying our VW = VU +UW equation by PQ.PR.PS/D, we get RS.PQ =
PS.QR +PR.QS. The product of the diagonals of a cyclic quadrilateral is equal
to the sum of products of opposite

edges.

You might be wondering: what if U wasn't between V and W ? Well, for
that to happen, Q would have to be further round the circle towards P than one
of R and S; at which point Q would no longer be the vertex opposite P, one of R
and S would; and *its* projection onto the tangent would be between the
other two. So, as long as we label the vertex opposite P as Q, its projection U
necessarily is between the projections of the other two vertices.

Note that Pythagoras's theorem is a special case of Ptolemy's: combine your right-angle triangle with its image under a half turn about the mid-point of its hypotenuse, which is the centre of its circumcircle; apply Ptolemy's theorem to the resulting rectangle.

Consider how the products in Ptolemy's theorem pair up the edges: each pair, that get multiplied together, uses all four corners; we pair an edge connecting two vertices of the quadrilateral with an edge connecting the other two. If we select one vertex of the quadrilateral, each multiplied pair has this vertex at one end of one edge; the other edge is an edge of the triangle among the remaining three vertices; and, within that triangle, it's the edge opposite the vertex to which the first edge connects our originally selected vertex.

In particular, if we connect each vertex of an equilateral triangle to any one point on its circumcircle, applying Ptolemy's theorem to the resulting quadrilateral would multiply each side of the triangle by one of the lines by which we connected its corners to the added point. Since the edges of the triangle are equal, we can cancel this common factor and Ptolemy's theorem reduces to the middle connecting line, which crosses a side of the equilateral triangle, being equal to the sum of the other two connecting lines.

So how about a regular pentagon ? That's got one vertex and side too many, so ignore one vertex and connect the two adjacent to it to complete a quadrilateral with the other two; the two diagonals of this cyclic quadrilateral now connect non-adjacent vertices of the pentagon – so these, like the edge we put in to cut off the surplus corner, are diagonals of the pentagon. The edges of the pentagon are all the same length, say p, and its diagonals are all the same length, call it d: so Ptolemy's theorem reduces to d.d = p.p +p.d and we can quickly infer that d/p is the golden ratio, 2.d = (1 +√5).p.

Written by Eddy.