Let $ABC$ be a triangle and let $O$ be the centre of its circumscribed circle. Points $X, Y$ which are neither of the points $A, B$ or $C$, lie on the circumscribed circle and are so that the angles $XOY$ and $BAC$ are equal (with the same orientation). Show that the orthocentre of the triangle that is formed by the lines $BY, CX$ and $XY$ is a fixed point.
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Tags: circumcircle, geometry
rafaello
27.02.2021 22:46
Trivial. Let $P=BY\cap CX$. Let $H$ be the midpoint of arc $BAC$, we claim that this is the orthocentre of triangle $PXY$. $$\measuredangle XHY+\measuredangle CXH=\frac{1}{2} \measuredangle BAC+\measuredangle CBH=90^\circ\implies PX\perp HY$$and $$\measuredangle HYB+\measuredangle XHY=\frac{1}{2} \measuredangle BAC+\measuredangle HCB=90^\circ\implies PY\perp HX.$$
Wictro
11.03.2021 22:30
Trivial. Let $P = BY \cap CX$ Let $H$ be the midpoint of arc $BAC$ - we claim that this is the orthocenter of triangle $PXY$. If we let $Q$ be the midpoint of arc $BC$, we clearly have $XY = QB = QC$ and, regardless of the configurations, we have $QY \parallel CX$ and $QX \parallel BY$. This means that $PXQY$ is a parallelogram. Since ${Q, H}$ are antipodal in $(ABC)$, our claim holds from "Reflecting the orthocenter Lemma".