Finding equilateral triangle such that two vertices are on rays of an angle and the other vertex is a fixed point inside that angle. Assume A is a fixed point inside an angle, How can one find two points B and C on different rays of that angle such that ABC will be an equilateral triangle?
I know how to analytically solve this problem but I want an elementary method by using 4 well known maps (translation, reflexion, rotation and scaling). 

Let us fix the notation. From the point $O$ we draw two rays $Ob$ and $Oc$, and let $A$ be a point inside the angle $\widehat {bOc}$.
From $A$ we rotate the ray $Oc$ with an angle of $\pm 60^\circ$, thus obtain intersection points $B_\pm$ on $Ob$ with the rotated rays. (We may possibly get points at infinity, or the one intersections intersects only the prolongation of $Ob$ in the opposite ray on the same line.)
We rotate back, the points $B_\pm$, get points $C_\pm$.
So we may get using this construction two solutions, $\Delta AB_-C_-$ and $AB_+C_+$.

Reciprocally, each solution $\Delta ABC$ equilateral with $B\in Ob$, $C_Oc$ is found by one of the above constructions, because we can rotate $C$ with $\pm 60^\circ$, and get $B\in Ob$.

A picture:

Here, $Oc$ was rotated with $\pm 60^\circ$ by rotating two humaly chosen points of $Oc$, in my choice, $\color{orange}{O}$, the origin of the ray $Oc$, and $\color{green}{\Pi}$, the projection of $A$ on $Oc$.
