This theorem is hardly ever studied in high-school math. So we will need to recall what the theorem actually says. Let us remember a simple fact about triangles. If ABC is a triangle then it is always possible to circumscribe this triangle in a circle (which is unique). In fact, we will prove this result later on in this post when we generalize Ptolemy's theorem.

Let us come up with some unusually strange terminology which is never used in any geometry text. Instead of saying "circumscribe" we will say "circumcise". If some shape (like a triangle or quadrilateral) can be circumcised i.e. it is possible to circumscribe it in a circle, we call call this figure "male". If some shape cannot be circumcised in a circle then we will call this figure "female".

If we have a quadrilateral (four-sided figure) ABCD (where the vertices, by convention, are named in cyclic order - this is never an issue with a triangle because the triangle has just three sides) then it is not always possible to circumcise this quadrilateral. To see why not suppose that it was possible to. Then ABC is circumcised in a unique circle. Since ABCD is assumed to have been circumcised in a circle the fourth vertex, D, must lie on this circle. And now just move this vertex around away from the circle. Since the circle circumcising ABC is unique it means the resulting quadrilateral with a moved vertex D is female.

Therefore, by the above observations, we note that all triangles are male. However, not all quadrilaterals are male. Some quadrilaterals are male and some are female. Ptolemy realized a simple way to test if a quadrilateral is male or female.

**Ptolemy's Theorem**: Let ABCD be a quadrilateral, this quadrilateral is male if and only if $[AC][BD]=[AB][DC]+[AD][BC]$.

In case, it is not clear, the meaning of $[AC]$ means the distance from A to C, and so on. Draw a picture for yourself of this theorem and it becomes very easy to see. Ptolemy's theorem says that a quadrilateral is male only if

*the sum of the product of opposite sides is equal to the product of the diagonals*.

There are many proofs to Ptolemy's theorem. Go to Wikipedia here. There are four proofs given to this theorem. One is geometric (classical), two trigonometric and algebraic (using complex numbers). These proofs are complicated! Look at how long and complicate they are. And it seems that these proofs do not even prove the converse! They just show that male quadrilaterals satisfiy the cross diagonal product formula, thereby only giving a necessary condition.

I have a different proof. A proof that is extremely easy and beautiful. And this proof proves the necessary and sufficient condition simultaenously. The smoother proof of Ptolemy's theorem uses "inversion in a circle" which can be found here. But the circle inversion proof is still geometric. You still need to draw pictures and calculate lengths.

I have a much easier proof of Ptolemy's theorem. This proof is nothing original. All I did was take the circle inversion proof and clean it up even more. I never seen anyone ever did this adaptive proof before, but I think this is by far the easiest proof of Ptolemy's theorem.

**Proof**: We will use Mobius transformations to prove Ptolemy's theorem. The only Mobius transfarmotion that we will use is "inversion" it is the mapping $M(z) = \frac{1}{z}$. Note we use the variable $z$ to denote that this function is a complex function it takes it values for complex numbers. The Mobius map $M$ has some really nice properties. The property that we want is the following. If $C$ is a circle passing through the origin then $M(C)$ (the image of the circle under the map $M$) is a line. Of course, $M$ is not defined at the origin, so if we want to be a little more formal we should say $M(C \setminus \{ 0 \})$ is a line. But I think this is clear from the context.

If $p$ is some point in the complex plane then $M_p(z) = \frac{1}{z+p}$ is basically the same as the Mobius map above it is just translated to a different point that acts as the new origin.

Let use denote the verticies of a male quadrilateral ABCD as $z_1,z_2,z_3,z_4$ in that order, these points will be the corresponding complex points representing the vertices of our quadrilateral. These points are assumed to lie on a circle. Consider the function $\frac{1}{z-z_1}$. This is a Mobius transformation like we have discussed above. And so it maps circles passing through the origin (in this case $z_1$) to lines. If C is a circle that circumcises ABCD then C is mapped to a line under the function $\frac{1}{z-z_1}$ because C passes through $z_1$. Since $z_2,z_3,z_4$ lie on this circle it means the corresponding points under this function $w_2=\frac{1}{z_2-z_1}$, $w_3=\frac{1}{z_3-z_1}$, and $w_4=\frac{1}{z_4-z_1}$ lie on a line. The point $w_3$ lies in between $w_2$ and $x_4$ because $z_3$ was the vertex in between $z_2$ and $z_4$ (this is why we are careful about the order in which we label the verticies). Since they lie on a line it clearly follows that that distance from $w_2$ to $w_3$ plus the distance from $w_3$ to $w_4$ is the same as the distance from $w_2$ to $w_4$:

$d(w_2,w_3) + d(w_3,w_4) = d(w_4 ,w_2)$

Therefore,

$d\left( \frac{1}{z_2-z_1} , \frac{1}{z_3 - z_1} \right) + d\left( \frac{1}{z_3 - z_1} , \frac{1}{z_4 - z_1} \right) = d\left( \frac{1}{z_4 - z_1} , \frac{1}{z_2 - z_1} \right)$

Combine,

$\frac{d(z_2 ,z_3)}{d(z_2,z_1)d(z_3,z_1)} + \frac{d(z_3,z_4)}{d(z_3,z_1)d(z_4,z_1)} = \frac{d(z_4,z_2)}{d(z_2,z_1)d(z_4,z_1)}$

Clear denominators,

$d(z_2,z_3)d(z_4,z_1) + d(z_3,z_4)d(z_2,z_1) = d(z_4,z_2)d(z_3,z_1)$.

This proves the theorem because $d(z_2,z_3)$ is just [BC], $d(z_4,z_1)$ is just [AD] and so forth. So we get that $[AC][BD]=[AB][DC]+[AD][BC]$. In fact, if you notice we can retrace all our steps backwards. All of these steps in the proof are "if and only if" statements, therefore by just following the proof backwards we get the converse statement.

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What we can do now is generalize Ptolemy's theorem. We can consider a polygon with n sides. The proof is identical with points called now $z_1,z_2,...,z_n$. By using the Mobius transformation $\frac{1}{z-z_1}$ it is clear what the situation is if it is written out algebraically:

$$\displaystyle \sum_{k=2}^{n-1}d(z_k , z_{k+1}) \prod_{j\not = k,k+1} d(z_j , z_1) = d(z_n , z_2) \prod_{j\not = 2,n} d(z_j , z_1)$$.

What this is saying geometrically is not so complicated. Fix a vertex, in this case $z_1$. Then consider two adjacent verticies (not containing $z_1$). Compute their product of their distance and the distances from $z_1$ to all other different vertices. Now sum this product over all such adjacent sides. This resulting number will be equal to the the product of the distances from $z_2$ and $z_n$ and all other vertices from $z_k$ to $z_1$ only if the polygon is male. And this gives us a method to determine which polygons are male.

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We can also prove Ptolemy's inequality. Which says that if ABCD is a quadrilateral then $[AC][BD]\leq [AB][DC]+[AD][BC]$. This proof is immediate from our previous work. If ABCD is a quadrilateral then fix a vertex, say $z_1$ and again consider the Mobius transformation $f(z) = \frac{1}{z-z_1}$. This image of the vertices B,C,D under this Mobius map will be mapped to three pair that is not necessarily a line because we are not assuming that B,C,D lie in a circle. Therefore, the distance from f(B) to f(C) plus distance from f(C) to f(D) is greater or equal to the distance from f(B) to f(D). This follows from the triangle inequality. Because since f(B),f(C),f(D) are not necessarily straight they form a circle. So the sum of their sides exceeds the length of the third side. Therefore, we always get $[AC][BD]\leq [AB][DC]+[AD][BC]$ for any quadrilateral. Equality holds only in the case if ABCD is male.

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