Prove of a theorem of a geometrical place I am having issues to prove the back of this theorem: 

Let $ABC$ be a triangle and fixed $D∈AB$. The Geometric Place of the $X$-points that form with $D$ and an arbitrary point $S∈AC$ an equilateral triangle $DSX$ is a straight segment.

Can someone save me?
 A: 
Note that $a$ and $\beta$ are constants, only $s$ and $\delta$ are changed. Also note that according to law of sines:
$$a\sin\beta=s\sin\delta\tag{1}$$
Let us calculate $x$ and $y$ coordinates of point $X$:
$$x=BS+s\cos(180^\circ-(\delta+60^\circ))=a\cos\beta+s\cos\delta+s\cos(120^\circ-\delta)$$
$$x=a\cos\beta+s\cos\delta+s\cos120^\circ\cos\delta+s\sin120^\circ\sin\delta$$
$$x=(a\cos\beta+s\frac{\sqrt3}{2}\sin\delta)+\frac12 s\cos\delta$$
Now use (1) and you get:
$$x=(a\cos\beta+a\frac{\sqrt3}{2}\sin\beta)+\frac12 s\cos\delta\tag{2}$$
On the other side:
$$y=s\sin(180^\circ-(\delta+60^\circ))=s\sin(\delta+60^\circ)$$
$$y=s\sin60^\circ\cos\delta+s\cos60^\circ\sin\delta$$
Now use (1) and you get:
$$y=\frac12 a\sin\beta+\frac{\sqrt3}{2}s\cos\delta\tag{3}$$
Introduce variable $u=s\cos\delta$ and take a closer look at (2) and (3). As already mentioned, $a$ and $\beta$ are constants so we can write (2) and (3) in the following form:
$$x=c_1+c_2u\tag{4}$$ 
$$x=c_3+c_4u\tag{5}$$
...with $c_1,c_2,c_3,c_4$ being constant values and $u$ being a parameter. Because of that equations (4) and (5) are actully parametric equations of a straight line (you can eliminate $u$ and write $y$ as a linear function of $x$ directly if you want so, but that's really not necessary, except if you want to investigate the locus of point X further).  
A: Most of the setup is irrelevant to the movement of $X$. $\Delta DSX$ being isosceles will suffice:

All we need is that $S,X$ are transformed by the same angle and the same scale of length relative to $D$. Then $\Delta DSS'$ and $\Delta DXX'$ are congruent. Here is a diagram. Details are left to you, but feel free to ask.

$\qquad\qquad\qquad$ 
A: It is well known that isometrie (i.e. rotation, reflection, translation, glide translation) takes line to a line, segment to a segment, circle to a circle. 
Since $S$ goes to $X$ after a rotation around $D$ for $60^{\circ}$, we can aplay this fact a mentioned and we are done.
Notice that $D$ does not need to be on a segment $AB$ and this is still true.
