Over the years I collected a few results in Euclidean geometry that struck my fancy. The proof presented here are sometimes my own work, and sometimes borrowed from other sources, but unfortunately from where often lost to my memory. Recently I had cause to resurrect some of them and redraw and recast the proofs for presentation.

All these results can be derived using only elementary techniques.

Given three circles each touching the other two, then it is possible to draw a fourth circle touching all the other three. Find the relation between the radii of these circles.

Since each of the circles touches all the other three, then the relation found will be symmetrical in each of the radii. Further it is clear that, if the first three all touch each other externally, then the fourth can be placed either so that it touches the others also externally, or that the given circles touch the fourth internally. This points to a quadratic relation, symmetric in the four variables, giving two solutions.

We can define the *bend* of a circle to be the reciprocal of its radius.
If the radii of each of the four circles are taken to be *r _{i}*,
for

There are two theorems named after Ptolemy, and they are closely related. Both refer to a convex cyclic quadrilateral and relate the lengths of the diagonals to the lengths of the sides. Usually only the first is shown and given the name Ptolemy's theorem.

The first expresses the product of the two diagonals to the sum of products of pairs of opposite sides. The second expresses the ratio of the diagonals as a ratio of sums of products of adjacent sides.

First theorem: *ef* = *ac* + *bd*

Second theorem:
*e* / *f* = (*ad* + *bc*) / (*ab* + *cd*)

This theorem was discoverd by the Japanese during their period of isolation from the Western world. It did not occur in Europe independently.

In a convex cyclic polygon of *n* sides, diagonals are drawn to create *n*-2
triangles. Inside each of these triangles the incircle is drawn. The sum of the radii
does not depend on how the triangulation is accomplished.

In the diagrams above, only a quadrilateral is shown, and it is only necessary to prove this case in detail, since it is possible to maniplate one triangulation into any other by swapping diagonals in quadrilaterals. In this case

*r _{A}* +

This result occurs as a theorem in the appendix to the book *Japanese Temple Geometry*
by Eiichi Ito et al, where it is reported without proof. It is given the uninspiring
name of “Theorem 9”.

Sangaku are tablets hung up in temples and shrines of Japan during the Edo, or Tokugawa, period when Japan had virtually no contcat with European developments in science and mathematics. The problems portrayed on these tablets vary from the trivial to very difficult, but there are some fairly common themes among them, in particular the use of circles inscribed in triangles. This result is such a one.

*n*+1 arbitrary points, labelled *P _{i}*, for

The radius of the incircle of triangle *ABC* is *R*, and the height is *h*.

Prove that 1 – (2*R*/*h*) =
(1–(2*r*_{1}/*h*)) (1–(2*r*_{2}/*h*)) ...
(1–(2*r*_{n}/*h*))

The art of paperfolding can do things that using the Euclidean straight edge and compass cannot. One of those things is finding the root of a cubic, and that ability enables the trisection of an angle into three equal parts.

So, the question is, how?

As a hint, the key is the fact that when folding a piece of paper it is possible to place two distinct known points onto two separate lines with one fold.

For any planar quadrilateral, construct squares on the same side of each side. Then the lines joining the centres of squares on opposite sides are equal in length and perpendicular to one another.

The diagram shows only the convex case, with squares drawn outside the quadrilateral.

To cater for the general case, where the quadrilateral could be concave or even crossed, or where the squares are drawn toward the interior of the quadrilateral, we need to be clear how the squares are drawn. The phrase “on the same side of each side” is to be taken to mean that as you travel from one vertex of the quadrilateral to the next along a side, doing a complete cycle of all sides, then the squares are always drawn either to the left as you go, or to the right as you go.

To prove this little theorem is an exercise in transformations, although there undoubtedly will be other ways of doing it.

In a triangle, two angle bisectors are drawn, as from *A* and *B* in the diagram.
Then it is easy to prove that if the triangle is isosceles with angle *A* equal to *B*
then the two bisectors *AK* and *BL* are equal.

The Steiner-Lehmus theorem claims that the converse is also true. That is that *AK* = *BL*
implies that *A* = *B*. This is harder to prove, although many proofs are now known.

The diagram shows only the internal bisectors, but the theorem is also true when the bisectors are drawn externally to the triangle.

Here is a note to try to explain why it is difficult