NMO Selection Tests

Let $H$ be the orthocentre of triangle $ABC$, let $N$ be the midpoint of the segment $AH$ and let $D$ the foot of the perpendicular from $B$ to $AC$. We first prove that $\angle BNF = 90^\circ$. Let $F'$ be the intersection of line $AC$ and the perpendicular at $N$ to $BN$. The points $B, N, D$, and $F'$ lie on the circle on diameter $BF$, so
$$\angle NBF' = \angle ADN = \angle DAN = 90^\circ - \angle BCA = \angle ABO,$$
where the second equality holds on account of $DN$ being median in the right triangle $ADH$. Since $\angle NBF' = \angle ABO$, lines $BN$ and $BO$ are isogonal with respect to angle $\angle ABF$. Similarly, lines $AO$ and $AN$ are isogonal with respect to angle $\angle BAC$,

$$\angle AF'O = \angle BF'N = \angle BDN = \angle NHD = \angle ACB.$$
Hence $OF' \parallel BC$, so $F \equiv F'$, showing that $\angle BNF = 90^\circ$.

Next, we prove that $N, F$ and $M$ are collinear. Since $F \equiv F'$, points $O$ and $N$ are isogonal conjugates in triangle $ABF$, so
$$\angle NFB = \angle AFO = \angle ACB.$$
This implies $NF$ is tangent to circle $BFC$ and hence $N, F, M$ are collinear, as desired.
Let $AB$ and $A'M$ cross at $E$. Note that $E$ is the reflection of $A$ across $B$. Since $AK = 4AB$, point $E$ is the midpoint of the segment $AK$. In triangle $AEH$, line $BN$ is a midline, so $BN \parallel EH$. As $BN$ and $NM$ are perpendicular, so are $EH$ and $NM$. Moreover, $NH \perp EM$, hence $H$ is the orthocentre of the triangle $ENM$, implying $MH \perp EN$. Finally, in triangle $AKH$, line $EN$ is a midline, so $EN \parallel KH$ and hence $MH \perp KH$; that is, $H$ lies on the circle on diameter $KM$, as required.

Alternative solution for $\angle BNF = 90^\circ$. We use the same notations as in the previous solution. Consider the foot $P$ of the perpendicular from $F$ to $BC$. Then
$$2AN = 2NH = AH = 2 \operatorname{dist}(O, BC) = 2 \operatorname{dist}(F, BC) = 2FP,$$
so the segments $NH$ and $FP$ are parallel and have equal lengths, implying that $NFPH$ is a parallelogram. Similarly, $ANPF$ is also a parallelogram. Therefore, $HB \perp NP$, and since $NH \perp BC$, we deduce that $H$ is the orthocentre of the triangle $BNP$. Consequently, $HP \perp BN$, and since $HP \parallel NF$, it follows that $\angle BNF = 90^\circ$.