Simson line in the regular 17-gon

Here is a synthetic solution. First of all a picture. The given $17$-gon has in my notation the vertices $0,1,2,\dots,16$. (Because a computer aided solution would have them labeled as $\zeta^k$ for $k=0,1,2,\dots, 16$, where $k$ is the primitive $17$.th root of unity, $\zeta =\exp\frac{2\pi i}{17}$.)

The construction first:

We build the Simson line of $\boxed 8$ w.r.t $\Delta 0,7,13$.

For this, we have to construct the perpendiculars from $\boxed 8$ to the sides of the triangle. These perpendiculars pass further through points of the $34$-gon having $0,1,2,\dots,16$ also as vertices. We denote the vertex of this $34$-gon between the points $k$ and $(k+1)$ by $kA$, where $A$ is only a suffix. Then:

• The perpendicular from $\boxed 8$ on $7,13$ passes through $3A$.
• The perpendicular from $\boxed 8$ on $13,0$ passes through $13A$.
• The perpendicular from $\boxed 8$ on $0,7$ passes through $7A$.
• The Simson line is then the line through $(8,3A)\cap(7,13)$, and $(8,13A)\cap(0,13)$, and $(8,7A)\cap(0,7)$. (These three points are colinear.)

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Try here to find the solution without reading the following...

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The solution:

Let $P,Q$ be the projections of $8$ on the sides $7,13$ and $7,0$ of the given triangle:

Then we have $$ \widehat{7,P,Q} = \widehat{7,8,Q} = \widehat{7,8,7A} = \widehat{7,13,7A} \ . $$ (We have used $7,P,8,Q$ concyclic, sice the angles in $P,Q$ are right, and the fact that $8,Q$ passes through $7A$, and the equality of the two angles against the arc $7,7A$.)

In particular, $(P,Q)\|(13,7A)$.

It remains to observe that $(13,7A)\perp(14,15)$, since the angle between them is half of the difference of the measures of the arcs - from $15$ to $7A$, and - from $13$ to $14$, and we get $$ \frac 12\cdot2\pi\cdot\frac 1{17}\left(9\frac 12-1\right)=\frac \pi 2\ . $$

$\square$

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Note:

From the figure we observe the coincidence that the diametral line $5,13A$ also passes through the point of intersection of the Simson line and $(14,15)$. How can we prove and use this fact? (Consider $(12,13)\cap(14,15)$, and search for the mirror points...)