# Bottema in Three Rotations

### Wojtek Wawrów

27 May, 2014

The following theorem and proof can be found in Waldermar Pompe's booklet Wokół obrotów (Theorem 3.5), and can be thought as of a generalization of a generalization of Bottema's theorem:

Given hexagon $ABCDEF$ (not necessarily convex) in which $AB=BC,$ $DC=DE,$ $FE=FA,$ let $\alpha,$ $\beta,$ $\gamma$ denote the internal angles of this hexagon at the vertices $B,$ $D,$ $F,$ respectively. Assume $\alpha +\beta +\gamma=360^{\circ}.$

Then at vertices $B,$ $D,$ $F,$ $\Delta BDF$ has angles $\displaystyle\frac{\alpha}{2},$ $\displaystyle\frac{\beta}{2},$ $\displaystyle\frac{\gamma}{2},$ respectively.

### Proof

Let us also denote the other internal angles of hexagon as above. Note that because $\alpha +\beta +\gamma=360^{\circ}$ we also have $x+y+z=360^{\circ}.$ Thus there are two possibilities: either none exactly one of angles $x,y,z$ is at least $180^{\circ}.$ Without loss of generality assume $z\ge 180^{\circ}$ ((if none of the angles is concave, the argument won't change). Rotate $\Delta BCD$ around $D$ through angle $\beta$ so that point $C$ moves to point $E.$ Let $P$ be the image of point $B$ under this rotation. We can easily verify that $\angle PEF=x$ (because $\angle PED=y$ and $x+y+z=360^{\circ}.$Since we also have $PE=BC=AB$ and $EF=AF,$ we get $\Delta FAB=\Delta FEB.$ So, if we rotate $\Delta FAB$ around $F$ through angle $\gamma,$ we will get triangle $\Delta FEP.$

It follows that $BD=DP,$ $BF=PF,$ $\angle BDP=\beta,$ $\angle BFP=\gamma.$

Since $DBFP$ is a kite, diagonal $DF$ bisects two of its angles, giving us $\displaystyle\angle BDF=\frac{\beta}{2}$ and $\displaystyle\angle BFD=\frac{\gamma}{2}.$ That $\displaystyle\angle DFB=\frac{\alpha}{2}$ follows from $\alpha +\beta +\gamma=360^{\circ}.$

Now, how can this extend the generalization of Bottema's theorem? We'll show how this implies the generalized Bottema: apply above theorem to the hexagon $C_{B}DC_{A}ACB.$

Because $\angle C_{B}DC_{A}=180^{\circ},$ $\angle C_{A}AC+\angle CBC_{B}=180^{\circ}$ and $C_{B}D=DC_{A},$ $C_{A}A=AC,$ $CB=BC_{B},$ this is exactly the setup of the above theorem. So we instantly get $\displaystyle\angle ADB=\frac{180^{\circ}}{2}=90^{\circ}.$ We also get that angles $DAB$ and $DBA$ do not depend on point $C$ (because angles $CAC_A$ and $CBC_B$ don't depend on it) so the angles of $\Delta ABD$ don't depend on $C.$ Thus the only technicality we are left with is that, given angles of a triangle and one of its sides, the triangle can be "flipped" in two ways. But we can rule out the possibility of point $D$ "leaping" from one position to the other by considering continuous movement of point $C.$

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