In the diagram, lines PS and PT are adjacent rays making angles ${\displaystyle \gamma _{n}}$  and ${\displaystyle \gamma _{n+1}}$  with line PR, which is perpendicular to the baseline, RST.

Line QXOY is parallel to the baseline and passes through O, the center of the incircle of ${\displaystyle \triangle }$  PST, which is tangent to the rays at W and Z. Also, line PQ has length ${\displaystyle h-r}$ , and line QR has length ${\displaystyle r}$ , the radius of the incircle.

Then ${\displaystyle \triangle }$  OWX is similar to ${\displaystyle \triangle }$  PQX and ${\displaystyle \triangle }$  OZY is similar to ${\displaystyle \triangle }$  PQY, and from XY = XO + OY we get

${\displaystyle (h-r)(\tan \gamma _{n+1}-\tan \gamma _{n})=r(\sec \gamma _{n}+\sec \gamma _{n+1}).}$ 

This relation on a set of angles, ${\displaystyle \{\gamma _{m}\}}$ , expresses the condition of equal incircles.

To prove the lemma, we set ${\displaystyle \tan \gamma _{n}=\sinh(a+nb)}$ , which gives ${\displaystyle \sec \gamma _{n}=\cosh(a+nb)}$ .

Using ${\displaystyle a+(n+1)b=(a+nb)+b}$ , we apply the addition rules for ${\displaystyle \sinh }$  and ${\displaystyle \cosh }$ , and verify that the equal incircles relation is satisfied by setting

${\displaystyle {\frac {r}{h-r}}=\tanh {\frac {b}{2}}.}$ 

This gives an expression for the parameter ${\displaystyle b}$  in terms of the geometric measures, ${\displaystyle h}$  and ${\displaystyle r}$ . With this definition of ${\displaystyle b}$  we then obtain an expression for the radii, ${\displaystyle r_{N}}$ , of the incircles formed by taking every Nth ray as the sides of the triangles

${\displaystyle {\frac {r_{N}}{h-r_{N}}}=\tanh {\frac {Nb}{2}}.}$ 

• Hyperbolic function
• Japanese theorem for cyclic polygons
• Japanese theorem for cyclic quadrilaterals
• Tangent lines to circles

• Equal Incircles Theorem at cut-the-knot
• J. Tabov. A note on the five-circle theorem. Mathematics Magazine 63 (1989), 2, 92–94.