Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 6.7 National olympiad Prove it China

A right triangle $ABC$ , with $\angle BAC = 90^\circ$ , is inscribed in the circle $\Gamma$ . The point $E$ lies in the interior of the arc $\text{BC}$ (not containing $A$ ), with $EA > EC$ . The point $F$ lies on the ray $EC$ with $\angle EAC = \angle CAF$ . The segment $BF$ meets $\Gamma$ again at $D$ (other than $B$ ). Let $O$ denote the circumcenter of the triangle $DEF$ . Prove that the points $A$ , $C$ , $O$ are collinear. (Posed by Bian Hongping)

Solution

Let $M$ and $N$ be the feet of the perpendiculars from $O$ to the lines $DF$ and $DE$ , respectively. Because $O$ is the circumcenter of the triangle $DEF$ , the triangles $EOD$ and $ODF$ are both isosceles with $EO = DO = FO$ . It follows that
$\angle EOF = \angle EOD + \angle DOF = 2\angle NOD + 2\angle DOM = 2\angle NOM.$
Because $\angle OND = \angle OMD = 90^\circ$ , the quadrilateral $OMDN$ is concyclic, from which it follows that $\angle NDM + \angle NOM = 180^\circ$ or $\angle BDN = \angle NOM$ . Because $ABED$ is concyclic, we have $\angle BAE = \angle BDE$ . Combining the above equations together, one has
$\angle EOF = 2\angle NOM = 2\angle BDN = 2\angle BDE = 2\angle BAE.$
Because $BC$ is a diameter of $\Gamma$ , it follows that
$\angle EOF + \angle EAF = 2\angle BAE + 2\angle EAC = 2\angle BAC = 180^\circ,$
from which it follows that $AEOF$ is concyclic. Let $\omega$ denote the circumcircle of $AEOF$ . Because $O$ lies on the perpendicular bisector of the segment $EF$ , $O$ is the midpoint of the arc $EF$ (on $\omega$ ), implying that $AO$ bisects $\angle EAF$ and $A$ , $C$ , $O$ are collinear.

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