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Croatian Mathematical Olympiad

Geometry Difficulty 6.4 National Olympiad Prove it Croatia

Let $\overline{AD}$ be the altitude of an acute-angled triangle $ABC$ . On the line $AD$ there are distinct points $E$ and $F$ such that $|DE| = |DF|$ and the point $E$ is inside the triangle $ABC$ . The circumcircle of the triangle $BEF$ meets segments $\overline{BC}$ and $\overline{AB}$ again at points $K$ and $M$ , respectively. The circumcircle of the triangle $CEF$ meets segments $\overline{BC}$ and $\overline{CA}$ again at points $L$ and $N$ , respectively.
Prove that the lines $AD$ , $KM$ and $LN$ are concurrent.

Solution

Notice that the circumcentre of triangle $BEF$ is on the segment $\overline{BC}$ . Therefore, segment $\overline{BK}$ is a diameter of the circumcircle of triangle $BEF$ , so $\angle BMK = 90^\circ$ , and analogously $\angle LNC = 90^\circ$ .

Let lines $AD$ and $KM$ intersect at $X$ . From $\angle BMX = \angle BMK = 90^\circ$ and $\angle XDB = \angle ADB = 90^\circ$ , it follows that the quadrilateral $BDXM$ is cyclic.

Quadrilaterals $BMEF$ and $EFCN$ are cyclic too, so from power of a point theorem (multiple use) it follows that
$|AX| \cdot |AD| = |AM| \cdot |AB| = |AE| \cdot |AF| = |AN| \cdot |AC|,$
and because of that, quadrilateral $CDXN$ is also cyclic.

Finally, $\angle XNC = 180^\circ - \angle CDX = 90^\circ = \angle LNC$ , i.e. point $X$ lies on the line $LN$ , which completes the proof.

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