An easier proof would be to just recall that the Simson line bisects the segment determined by the orthocenter of the triangle and the point, thus if we let X, Y be the midpoints of HP and HQ (where H is the orthocenter of ABC), the result follows at once by Desargues' theorem applied to triangles PXM, QYN, where M, N are the projections of P, Q on AB.

Best,

Cosmin

--- In Hyacinthos@yahoogroups.com, "Alexey Zaslavsky" <zasl@...> wrote:

>

> Dear colleagues!

> Is next fact known?

> Let P, Q be two point on the circumcircle of triangle ABC. Then their Simson lines meet on the altitude from C iff PQ is parallel to AB.

>

> Sincerely Alexey

>

> [Non-text portions of this message have been removed]

>