SAUDI ARABIAN IMO Booklet, 2023

On $MN$, take the point $D$ such that $ND = AM$. Then, $AN = DM$. According to the sine theorem then
$$\frac{BM}{\sin \angle BNM} = \frac{BN}{\sin \angle BMN} = \frac{AC}{\sin \angle AMC} = \frac{MC}{\sin \angle MAC},$$

so $\angle BNM = \angle MAC < 90^\circ$, which implies that $\triangle AMC \cong \triangle NDB$ so
$$\angle BDM = 180^\circ - \angle BDN = 180^\circ - \angle AMC = \angle AMB = 60^\circ.$$
Also, $BD = MC = MB$ so triangle $BDM$ is isosceles, with $\angle BDM = 60^\circ$ so equilateral. Therefore $AN = DM = BM = \frac{1}{2}BC$.

Let $X, Y$ be the centers of the circumcircles of triangles $AMC, BMN$ and $H, K$ respectively their orthocenters. We have a familiar result: triangle $ABC$ has circumradius $R$ and orthocenter $H$ then $AH = 2R|\cos A|$. Applying to this problem, notice that two triangles $AMC$ and $BMN$ have the same radius of circumcircle $R$, so
$$MH = 2R \cos 60^\circ = R \text{ and } MK = 2R |\cos 120^\circ| = R.$$
Therefore, four points $X, Y, H, K$ are on the same circle with center $M$. Next, denote $\ell$ as the internal bisector of $\angle BMN$, then $MH, MX$ are symmetric through $\ell$. And $MH = MX$ so we have $HX \perp \ell$. Similarly $KY \perp \ell$ so $HX \parallel KY$. From this it follows that $HXYK$ is an isosceles trapezoid. $\square$