Problem 2

Let $ABC$ be an acute triangle such that $AH = HD$, where $H$ is the orthocenter of $ABC$ and $D \in BC$ is the foot of the altitude from the vertex $A$. Let $\ell$ denote the line through $H$ which is tangent to the circumcircle of the triangle $BHC$. Let $S$ and $T$ be the intersection points of $\ell$ with $AB$ and $AC$, respectively. Denote the midpoints of $BH$ and $CH$ by $M$ and $N$, respectively. Prove that the lines $SM$ and $TN$ are parallel.

 

Solution

Let L be the midpoint of AC, O be the midpoint of AB, E, F the traces of the heights from A, B respectively.



It is  \angle{ATH} = 90^{\circ} - \angle{EHT} = 90^{\circ} - \angle{SHB} = 90^{\circ} - \angle{HCB} = \angle{B},


  А quadrilateral BCTS is cyclic , so \angle{ATS} = \angle{B}.

Now LN \parallel BH \leftrightarrow \angle{LNH}=\angle{AHF}=\angle{B}=\angle{ATS}.

А quadrilateral THNL  is cyclic , so  \angle{HTN}=\angle{HLN}=90^{\circ} we get that HL \parallel BC.

We also show that   \angle{HSM}=90^{\circ}   and we are done.

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