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Geometry Difficulty 8.5 Shortlist Prove it Baltic Way

Let $\triangle ABC$ be an acute triangle. Denote by $E$ and $F$ the feet of the altitudes from $B$ and $C$ , respectively. Let $H$ be the intersection of $BE$ and $CF$ . Let $D$ be on the same side of line $BC$ as $A$ and satisfy:
$\angle DBC = \angle DCB = \angle BAC.$
Let $N$ be the midpoint of $EF$ . Prove that points $H$ , $D$ and $N$ are collinear.

Figure 23

Solution

Refer to figure 23. Notice that as triangles $HEF$ and $HCB$ are similar with different orientations and $C$ , $H$ , $F$ and $B$ , $H$ , $E$ collinear, then the statement is equivalent to $DH$ being a symmedian from $H$ in $BHC$ .

As $\angle CBD = \angle BAC = 180^\circ - \angle BHC$ , the line $DB$ is tangent to $o(BHC)$ . Similarly, $DC$ is tangent to $o(BHC)$ . The line $HD$ is the line connecting $H$ with the intersection of tangents to $o(BHC)$ at $B$ and $C$ , and so $HD$ is indeed the symmedian in $BHC$ .

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.