Gauss-Bodenmiller Theorem

The Gauss-Bodenmiller Theorem is a theorem in geometry concerning complete quadrilaterals and the circles on its diagonals.


Statement

Consider quadrilateral $ABCD$ with $P=AD \cap BC$ and $Q = AB \cap CD$. Then, the circles with diameters $\overline{AC}$, $\overline{BD}$, and $\overline{PQ}$ are coaxial. Their radical axis is a line passing through each of the four orthocenters of the triangles $PAB$, $PCD$, $QAD$, and $QBC$.


Proof Sketch

The theorem looks complicated, but the proof is quite simple. The idea is to just take every orthocenter and proves that it has the same power with respect to all three circles. Then, the orthocenters lie on the radical axes, which implies the conclusion.

Proof

Let $\omega_1$, $\omega_2$, $\omega_3$ denote the circles with diameters $\overline{PQ}$, $\overline{AC}$, $\overline{BD}$, respectively.

Let $H_1$ denote the orthocenter of triangle $BCQ$. It is not hard to see that $H_1$ is the radical center of $\omega_1$, $\omega_2$ and the circle with diameter $\overline{QC}$: note that $Q$, $C$, the foot of $Q$-altitude, foot of $C$-altitude are concyclic with diamter $QC$.

Also, $Q$ and foot of $Q$-altitude lie on $\omega_1$, and $C$ and foot of $C$-altitude lie on $\omega_2$. Therefore, $H_1$ lies on the radical axis of $\omega_1$ and $\omega_2$. Similarly, $H_1$ lies on the radical axes of all three pairs, so it is the radical center.

Similarly, the other three orthocenters also have the same properties, and this is only possible if the radical axes of all three pairs coincide, as desired. Therefore, all four orthocenters are collinear, and they lie on the so called Steiner line.

Note that this also proves the existence of the Gauss line, which is defined as the line passing through the midpoints of $\overline{AC}$, $\overline{BD}$, and $\overline{PQ}$, which are just the centers of the three circles, which lies perpendicular to the Steiner Line.

See Also

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