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Given a scalene triangle $\triangle ABC$ with $H$ the orthocenter of the triangle. The internal bisector of the angle $\angle BAC$ intersects the lines $BH$ and $CH$ at the points $Λ$ and $Θ$ correspondingly. If (c) is the circumcircle of the triangle $\triangle ΗΛΘ$ and (e) is the tangent of (c) at $Λ$. The perpendicular line from $H$ towards (e) is the line (e) at the point $N$. If $T$ is the point where the perpendicular line from $Λ$ towards $ΗΘ$ intersects the line $ΗΘ$, prove that $NT$ and $ΛΘ$ are parallel.

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My thoughts are the following:

$\angle ΛΤΗ+\angle ΗΝΛ=90^{o}$

Hence $ΤΗΝΛ$ is inscribable.

Hence we have that $\angle ΤΛΝ+ \angle ΤΗΝ=180^{o}$

And also $\angle ΛΝΤ=\angle ΛΗΤ$

Hence it is enough to prove that $\angle ΛΗΘ = \angle ΛΘΗ$.

Moreover it seems like $\triangle ΛΘΗ$ is equilateral, but I can't prove that. Could you please explain to me how to solve this question?

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1 Answer 1

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I will call the points where the angle bisector of $\angle BAC$ intersects $BH$ and $CH$ as $P$ and $Q$ respectively. Let $S$ be some arbitrary point on the right[Since $P$, $Q$ are on the right of $H$ ] of point $P$ on the tangent to the circumcircle of $\triangle HPQ$.

Observe that, $\angle SPQ=\angle PHQ=\angle PHT=\angle PNT=\angle SNT$ and thereafter $PQ\parallel NT$.

$\triangle HPQ$ is not necessarily equilateral but isosceles since $\angle HPQ=\angle HQP=90-\frac {\angle A}{2}$.

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