Let $P$ be midpoint of $AH$. Then $P,N,M$ lie on the Newton Gauss line of quadrilateral $BB_1CC_1$. Also, by shooting lemma (since $PB_1MC_1$ is cyclic with $PB_1=PC_1$), we get $PB_1^2=PN \cdot PM$. Combined with $PH=PB_1$, we have the result.

Let $U$ be midpoint of $AH$ then: $UH^2 = UB^2_1 = UC^2_1 = \overline{UM} . \overline{UN}$ of $AH$ tangents $(HMN)$ at $H$

Let $AA_1$ the third altitude. We now that $M$ is the circumcenter of $BC_1B_1C$. Then $MN \perp B_1C_1$ at $N$. $$\angle HNM = 90- \angle C_1NH$$ $$\angle MHA_1 = 90- \angle HMB$$ Since $\triangle HB_1C_1$ is similar to $\triangle HCB$ we have $\angle HMB = \angle HNC_1 \Rightarrow \angle HNM= \angle MHA_1$

Gauss theorem for $BB_{1}CC_{1}$ $M$,$N$,$\frac{AH}{2}$ collinear. $\frac{AH}{2}=K$ $MC_1=MB_1=\frac{BC}{2}$ $MN$ perpendicular $B_1C_1$ $AH$ and $B_1C_1$ intersect at $P$ $(A,H,P,A_1)=-1$ $KP \cdot KA_1=KH^2=KN \cdot KM$

ISL 2009 G4 wrote: Given a cyclic quadrilateral $ABCD$, let the diagonals $AC$ and $BD$ meet at $E$ and the lines $AD$ and $BC$ meet at $F$. The midpoints of $AB$ and $CD$ are $G$ and $H$, respectively. Show that $EF$ is tangent at $E$ to the circle through the points $E$, $G$ and $H$. The end $\blacksquare$ This config was also present in USA TSTST 2012 #7

Firstly we note that <C₁CB=α and <B₁BC=β.We get <CAH=β=<B₁C₁H; <C₁AH=α=<C₁B₁H; <C₁BB₁=<C₁CB₁=90-α-β.HN and HM is median and we get with median sine <C₁HN=<BHM.AH intersect BC at D.We note that <DHM=θ.We get <HMB=90-θ,<AHN=α-β+θ.We want to find <HNM=θ.It is easy. Because MC₁=MB₁=MB=MC .We are done.

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