In geometry, given a triangle $ABC$ and a point $P$ on its circumcircle, the three closest points to $P$ on lines $AB$, $AC$, and $BC$ are collinear. The line through these points is the Simson line of $P$, named for Robert Simson.
Let $A_1A_2\ldots A_{17}$ be a regular 17-gon as in the following figure:
Question: How can one prove that the Simson line of $A_9$ with respect to the triangle $A_1A_8A_{14}$ is perpendicular to the side $A_{15}A_{16}$?
On
This result generalizes nicely, and leads to a very neat follow-on result.
Abusing notation to define $\operatorname{cis}\theta := (\cos\theta, \sin\theta)$, we can coordinatize with $A_m := \operatorname{cis}2\theta_m$.
The Simson line of $\triangle A_i A_j A_k$ for $A_p$ has direction vector $$\operatorname{cis}(\theta_i+\theta_j+\theta_k-\theta_p) \tag{1}$$
The line $\overleftrightarrow{A_qA_r}$ (with $\theta_q\neq \theta_r$) has direction vector $$\operatorname{cis}\left(\frac{\pi}{2}+\theta_q+\theta_r)\right) \tag{2}$$
These vectors ---and therefore the corresponding lines--- are perpendicular if and only if $$\sin(\theta_i+\theta_j+\theta_k-\theta_p-\theta_q-\theta_r) = 0 \tag{3}$$ That is,
$$\theta_i+\theta_j+\theta_k = \theta_p+\theta_q+\theta_r \pmod{\pi} \tag{$\star$}$$
For the points $A_m$ on a regular $n$-gon, we can take $\theta_m = \pi m/n$ (recall that we incorporated a $2$ into the definition of $A_m$, so we don't use $2\pi$ here). In this case, $(\star)$ becomes $$i+j+k = p + q + r \pmod n \tag{4}$$
For the problem at hand, we have $n=17$, $i=1$, $j=8$, $k=14$, $p=9$, $q=15$, $r=16$. And, indeed, $$i+j+k = 23 = 40 = p+q+r \pmod{17}$$ This verifies the claim. $\square$
But there's more to say ...
Given that $p$, $q$, $r$ are interchangeable in $(\star)$, we have a situation where the Simson line of $\triangle A_i A_j A_k$ with respect to any of $A_p$, $A_q$, $A_r$ is perpendicular to the line through the other two. But, the sets $\{i,j,k\}$ and $\{p,q,r\}$ are also interchangeable: the Simson line of $\triangle A_p A_q A_r$ with respect to any of $A_i$, $A_j$, $A_k$ is perpendicular to the line through the other two.
We could say, then, that $\triangle A_i A_j A_k$ and $\triangle A_p A_q A_r$ are "ortho-Simsonian conjugates".
(Actually, "ortho-Simsonian" seems to suggest that there's some orthogonality among Simson lines, so a better term is needed.)
Here's a diagram of OP's scenario, with $\triangle A_1 A_8 A_{14}$ and $\triangle A_9 A_{15} A_{16}$.
We see that, as a bonus, all six Simson lines are concurrent! The point of concurrency is
$$\frac12\left(A_i+A_j+A_k+A_p+A_q+A_r\right) \tag{$\star\star$}$$
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:
Try here to find the solution without reading the following...
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 $$ \frac 12\cdot2\pi\cdot\frac 1{17}\left(9\frac 12-1\right)=\frac \pi 2\ . $$
$\square$
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...)