A geometry or trigonometry problem from high school... I found this problem from some old notes on high school geometry and tried solving it. I have been able to "prove" this by setting up a coordinate system with a circle of unit radius and then using brute force algebra with the coordinates, which was rather tedious. I am curious to see if this can be solved using Euclidean geometry or trigonometry or even complex numbers on the unit circle in a simpler way. I present the problem and the solution below.
Problem: Consider a circle with a fixed radius. Let A, T and S be three points on the circumference of the circle such that ATS is an acute-angled triangle. Let E and F be two points on the sides AS and AT, respectively, such that the segment EF is parallel to the segment TS. Let M be the midpoint of EF and let MN be the perpendicular bisector of EF such that it meets TS at the point N. Extend MN to the point K such that N is the midpoint of MK. (Thus, MN=NK in length and TS is the perpendicular bisector of MK.) If the point K also lies on the circumference of the circle (i.e., ATSK is a cyclic quadrilateral), then prove that the angle KFE (acute angle made by segments KF and FE) is equal to the angle TAS (angle of the triangle ATS made by segments TA and AS).
Solution: Let us consider a circle of unit radius and set up a coordinate system with the origin at the center of the circle $O=(0,0)$. Orient the circle so that the y-axis is perpendicular to the chord $TS$. Thus, the y-axis bisects the chord $TS$ and $T$ and $S$ are reflections of each other through the y-axis. Let $\vec{OS} = (t,-s)$ give the coordinates of the point $S$ and, therefore, $\vec{OT} = (-t,-s)$. Thus, $T$ is to the left of the y-axis and $S$ is to the right. From this setup, we find that $\sin(\angle TAS) = |\vec{TS}|/2 = t$ (using law of sines for circumscribed circle). I will now show that $\sin(\angle KFE) = t$.
Let $\vec{OA} = (a_1, a_2)$ be the point $A$ on the circle. From the description of the problem, the points $F$ and $E$ lie on the chords $AT$ and $AS$, respectively. Therefore,
$$
\vec{OF} = \alpha\vec{OA} + (1-\alpha)\vec{OT}\; \mbox{ and }\; \vec{OE} = \alpha\vec{OA} + (1-\alpha)\vec{OE}\;,
$$
where $\alpha$ is some real number between $0$ and $1$. The point $M$ is the midpoint of $FE$ so $\vec{OM} = (\vec{OF}+\vec{OE})/2 = (\alpha a_1, \alpha a_2 - (1-\alpha)s)$. From the construction of the point $K$, it has the same x-coordinate as $M$, so let us write $\vec{OK} = (\alpha a_1, -b)$.
Using the fact that $|\vec{MN}| = |\vec{NK}| = b-s$ we obtain that the y-coordinate of $M$ is $b-2s$, which yields $\alpha = (b-s)/(a_2+2s)$. Also note that $\displaystyle \tan(\angle KFE) = \frac{|\vec{KM}|}{|\vec{FM}|} = \frac{2(b-s)}{(1-\alpha)t}$. I will show that this expression is in fact equal to $t/s = t/\sqrt{1-t^2}$ so $\sin(\angle KFE) = t$, which will complete the proof.
Since $K$ lies on the unit circle, we have $\alpha^2a_1^2 + b^2 = 1$. Substituting $\alpha$ and simplifying, gives us the following quadratic equation:
$$
(1+u^2)b^2 - (2su^2)b + (u^2s^2-1) =0\;, 
$$
where $u = a_1/(a_2+s)$. Solving for $b$ and simplifying (a bit tedious but not too bad---keep in mind $a_1^2 + a_2^2 =1$ and $t^2 + s^2=1$) and then substituting the solution for $b$ in $\alpha = (b-s)/(a_2+s)$ and further simplifying produces the following:
$$
b = s + \frac{t^2}{(1+u^2)(a_2 + s)}\; \mbox{ and }\; 1-\alpha = \frac{2s}{(1+u^2)(a_2+s)}\;.
$$
From this, we directly obtain: $\displaystyle \tan(\angle KFE) = \frac{2(b-s)}{(1-\alpha)t} = \frac{t}{s} = \frac{t}{\sqrt{1-t^2}}$. Therefore,
$$
\sin (\angle KFE) = t = \sin(\angle KFE)\;.
$$
This completes the "proof" using brute force algebra with coordinates.
I am wondering if a simpler coordinate-free proof may be available using, perhaps, smart use of Euclidean geometry.
 A: Here is an answer (partial), completely revised. The attached sketch is consistent with your conditions, and has certain additional parts described here.
Let $EF$ be produced in both directions to intersect the circle at $P$ and $Q$. Let $KF$ be drawn and produced to meet the circle again at $R$. Draw chord $SMR$.
Here is the part I am leaving out. It must still be proved that points $S$, $M$, and $R$ are collinear. I was able to do so, but it became so convoluted that I am sure there must be a better way. Let me leave that to you.

All arc measures here are central angles, not arc lengths.
$\stackrel{\LARGE\frown}{PS} + \stackrel{\LARGE\frown}{QR} = 2\angle PMS$
= $2\angle MST$ ... (alternate angles, $QP || TS$)
= $2\angle KST$ ... (reflection symmetry in kite $SMTK$)
= $\stackrel{\LARGE\frown}{KT}$
$\stackrel{\LARGE\frown}{PS} + \stackrel{\LARGE\frown}{QR} = \stackrel{\LARGE\frown}{KT}$
$\stackrel{\LARGE\frown}{PS} + \stackrel{\LARGE\frown}{SK}+ \stackrel{\LARGE\frown}{QR} = \stackrel{\LARGE\frown}{SK} + \stackrel{\LARGE\frown}{KT}$
$2\angle KFE = 2\angle TAS$
$\angle KFE = \angle TAS$
