Let
be the circumcenter and
the orthocenter of an acute-angled triangle
such that
. Let
be the foot of the altitude
of triangle
. The perpendicular to the line
at the point
intersects the line
at
. Prove that
.
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Let $O$ be the circumcenter and $H$ the orthocenter of an acute-angled triangle $ABC$ such that $BC>CA$. Let $F$ be the foot of the altitude $CH$ of triangle $ABC$. The perpendicular to the line $OF$ at the point $F$ intersects the line $AC$ at $P$. Prove that $\measuredangle FHP=\measuredangle BAC$.
Školjka