How exactly does multiplication make sense in synthetic geometry? I'll use a theorem expressing circle inversion. Let $C$ be some circle with radius $r$ and center $O$, and let $P'$ be some point outside $C$, then it has two tangents with $C$, which we will use to form the lines $QP'$ and $RP'$. Connect $R$ and $Q$ to one another and connect them to the center $O$. Then, $OQP'$ is a square triangle and is similar to the triangle $OQP$ by virtue of having the same angles. Therefore, $\frac{OP}{OQ} = \frac{OQ}{OP'}$.

What I don't get is, how are we justified from this last relationship to say that $OP*OP' = OQ^2 = r^2$? Doesn't this require multiplication, which is an algebraic property not available in synthetic geometry? $\frac{OP}{OQ} = \frac{OQ}{OP'}$ expresses nothing else than $OP$ is to $OQ$ like $OQ$ is to $OP'$, from which I wouldn't know how to derive something like multiplication, so I think I'm misunderstanding something here.

Here is a picture of the above I found on another thread Circle Inversion enter image description here

  • 1
    $\begingroup$ A possible way forward is that $OP' * OP$ actually defines a rectangle, which has the same area as the square $OQ * OQ$, but I'm trying to see whether this can be derived from $\frac{OP}{OQ} = \frac{OQ}{OP'}$. $\endgroup$
    – shintuku
    Mar 30, 2021 at 22:40
  • $\begingroup$ discovered useful-looking property: the circle with center situated halfway between $O$ and $P'$ and which has $OP'$ as a diameter intersects $C$ at $Q$ and $R$ $\endgroup$
    – shintuku
    Mar 30, 2021 at 23:07

2 Answers 2


It is possible to define proportionality relationships between line segments in purely geometric terms. The trick is that you force the line segments to be legs of a right triangle, which makes everything well-defined.

More formally, we say that $AB:AC=AD:AE$ if in the following diagram the hypotenuse line segments $BC$ and $DE$ are parallel:

right triangles

If $p,q,r,s$ are arbitrary line segments, then we can build a diagram like the above one using segments congruent to $p,q,r,s$ as legs of the two right triangles. We then say that $p:q=r:s$ if that diagram has parallel hypotenuses.

You can then define multiplication purely geometrically as well: you choose some particular line segment $1$, and for any segments $p$ and $q$ you say that the line segment $pq$ has length such that $1:q=p:pq$.

Once you have these definition, you can prove that in fact the definition $p:q=r:s$ has all the standard proportionality properties (e.g., that two triangles are similar if and only if their corresponding sides are proportional), and that the above multiplication forms a field along with the obvious addition operation on segment lengths by concatenation. These proofs are complicated and a little tedious but basically boil down to applying the inscribed angle theorem for circles lots of times in clever ways. You can find details in Hilbert's original book (though the Gutenberg version is kind of error-prone, so check his proofs carefully) or in Hartshorne's Euclidean geometry book.

  • $\begingroup$ ah... this is fantastic! it makes so many things much simpler! $\endgroup$
    – shintuku
    Mar 31, 2021 at 21:42

EDIT: Found a simpler, more intuitive proof:

enter image description here enter image description here

The parallelogram has twice the area of the triangle since they share the same base and are on the same parallels. This parallelogram shares the same parallels as the parallelogram $OP * OP'$, so they have the same area. I have the same conclusions on synthetic geometry multiplication in my old post below.

This is simply and application of Euclid Bk 1 P42 (https://mathcs.clarku.edu/~djoyce/elements/bookI/propI42.html)

Solved! Using part of Euclid's geometrical proof of the pythagorean theorem: https://mathcs.clarku.edu/~djoyce/elements/bookI/propI47.html

Using proof by construction, here's the proof that $OP * OP' = OQ^2$.

enter image description here

This means there is at least this analogy to multiplication in synthetic geometry, and it does depend on $\frac{OP}{OQ} = \frac{OQ}{OP'}$ but this actually gives us very little information on how to perform the 'multiplication': you actually have to construct said multiplication somehow.

If someone has alternatives, or more information on the nature of such constructions, it would be super appreciated! Another potential candidate for this proof would be a variation of Euclid's Book 6 Proposition 36: I'm not sure of this, but the constructions look similar.


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