# A Generalization of Van Aubel's Theorem

### Problem

The applet purports to illustrate the following statement:

If one erects right-hand ears on the sides of the quadrilateral $(z_0,\,z_1,\,z_2,\,z_3)\;$ with ear angles $\alpha$ at $z_0,\;$ $\beta$ at $z_1,\;$ $\frac\pi2-\alpha\;$ at $z_2,\;$ and $\frac\pi2-\beta\;$ at $z_3,\;$ the segments joining the opposite ear apices are perpendicular.

### Proof 1

The triangle $(0,\,1,\,Z(\alpha,\,\beta))\;$ of the upper half plane with angles $\alpha\;$ at $1\;$ and $\beta\;$ at $0\;$ is given by

$\displaystyle Z(\alpha,\,\beta)=\frac{\tan\alpha}{\tan\alpha+\tan\beta}(1 +i\tan\beta).$

The apex of the ear erected on the side $z_0\rightarrow z_1\;$ is $z_1 + (z_0 - z_1)\cdot Z(\alpha,\beta)\;$. The quotient of the vectors joining the opposite ear apices is thus

$\displaystyle\frac{z_3 + (z_2 - z_3)\cdot Z(\pi/2 -\alpha,\,\pi/2 -\beta) - \left[z_1 + (z_0 - z_1)\cdot Z(\alpha,\,\beta)\right]}{z_0 + (z_3 - z_0)\cdot Z(\pi/2 -\beta,\,\alpha) - \left[z_2 + (z_1 - z_2)\cdot Z(\beta, \,\pi/2 -\alpha)\right]},$

which simplifies to

$\displaystyle-i\frac{1+\tan\alpha\tan\beta}{\tan\alpha+\tan\beta}.$

### Remark

Observe that the final formula,

$\displaystyle-i\frac{1+\tan\alpha\tan\beta}{\tan\alpha+\tan\beta}$

is independent of the quadrilateral and, for $\alpha=\beta=45^{\circ},\;$ the ratio is $-i\;$ and its absolute value $1,\;$ as expected. This is also true for the left-hand ears, with the analogous formula obtained when the triangle $(0,\,1,\,Z(\alpha,\,\beta))\;$ is replaced with $(0,\,1,\,\overline{Z}(\alpha,\,\beta)).\;$

### Proof 2

For the basis quadrilaterals $(1,0,0,0),\;$ $(0,1,0,0),\;$ $(0,0,1,0),\;$ and $(0,0,0,1)\;$ it is easy to see with a figure and the sine law that the vectors joining the opposite ear apices are always at right angles with the same orientation and with length ratio $\displaystyle\frac{\cos (\alpha-\beta)}{\sin(\alpha+\beta)}.\;$ As these vectors are linear functions of the vertices, the vectors joining the opposite ear apices for any quadrilateral are also at right angles with the same orientation and length ratio.

### Acknowledgment

The statement is by Oai Thanh Dào; both proofs and the remark are due to Grégoire Nicollier.