Given three lines that intersect at a point, construct an equilateral triangle of given length so that its vertices lie on these lines.

Question

This problem has to be solved with a compass and a straight edge. I haven't found any solutions online.

Answer 1

We use the fact that a compass and a straight edge suffice for rotating any point $60^\circ $ about an origin.

Let $O$ be the point of intersection of the three lines. Choose an arbitrary point $A $ on $t $ on the right from $O$ and rotate the line $r $ $60^\circ $ clockwise about the point. Let the rotated line be $r'$. Let $B$ be the intersection point of $r'$ with $s$. Rotate the point $B $ $60^\circ $ counterclockwise, name the rotated point $C $. Observe that it lies on the line $r $. By the construction the triangle $ABC $ is equilateral and its vertices lie on the lines $r,s,t $. It remains only to rescale the triangle using homothetic transformation with respect to $O$.

Can you take it from here?

Note that the first point ($A$) can be chosen lying on any of three lines and the first rotation can be done both clock- and counterclockwise. The latter can be required for example if the angle between the lines $r$ and $s$ is $60^\circ $.