feet of the altitudes from H1 and lie on the Wallace-Simson line of H1. Since H1
is the circumcircle point of KΔ,P this line passes through Z, [2] Theorem 6. The
Wallace-Simson lines of H1 and P are perpendicular since these these points are
antipodal.
Theorem 16. If P is on the circumcircle of Δ, then the midpoints of Δ and Δ2 lie
on line L.
Proof. From point A1 the midpoints to A and A2 are on the line L1. Thus the
midpoint mA of A and A2 lies on a line parallel to L1. Since Δ and Δ2 lie on the
circumcircle centered at O the perpendicular bisector of A and A2 passes through
O and is perpendicular to L1. Thus mA lies on L. The argument is similar for the
other pairs of points and hence the desired result follows..
4. Equilateral triangles on equilateral hyperbolas
Proposition 17. Suppose ΔABC is an equilateral triangle on the right hyperbola
K. The circumcircle meets K at the fourth intersection point S. The center of K,
Z, is the midpoint of OS where O is the circumcenter of Δ.
Proof. Since Δ is equilateral H = O and the result follows since Z is the midpoint
of HS [2].