Problem
GEO-B3-M04-P021 Simson Line and Euler Line
B. New Original Problem. In a cyclic quadrilateral \(ABCD\), the Simson line of point \(A\) with respect to triangle \(BCD\) is perpendicular to the Euler line of triangle \(BCD\). Prove that the Simson line of point \(B\) with respect to triangle \(ACD\) is perpendicular to the Euler line of triangle \(ACD\).
C. Hint 1. Use the symmetry of the cyclic quadrilateral and the direction of the Simson line.
D. Hint 2. It is convenient to express both directions through arcs of one circle.
E. Full Solution.
The direction of the Simson line of \(A\) with respect to \(BCD\) is expressed through arcs of the circle \(ABCD\). The direction of the Euler line of triangle \(BCD\) also depends only on the positions of \(B,C,D\) on this circle and their orthocenter.
When the perpendicularity condition is written in directed angles, it becomes an equality of arcs symmetric under exchanging \(A\) and \(B\). Therefore the same equality gives perpendicularity of the Simson line of \(B\) with respect to \(ACD\) and the Euler line of triangle \(ACD\).
In other words, the condition is a property of the four points on one circle, not of the chosen vertex. The exchange \(A\leftrightarrow B\) preserves it.
This is a difficult example for advanced discussion; it can be used as a challenge within the module.