# 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?

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).

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.

$$\qquad\qquad\qquad$$

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.