how to construct an equilateral triangle whose one vertex is given and other two vertices lie on two parallel lines a vertex  and two parallel lines are given. construct an equilateral triangle whose other vertices lie on two parallel lines this is a question from the app euclidia. i tried to make equilateral triangle but one vertex is not touching the parallel line. please provide a solutionand how the construction works

 A: Let $A$ be a given point and $a$ and $b$ be given lines.
Also, let $R_{A}^{\alpha}$ be a rotation around $A$ by $\alpha$.
Now, let $$R_{A}^{60^{\circ}}(a)\cap b=\{C\}$$ and let $$R_{A}^{-60^{\circ}}(\{C\})=\{B\}.$$ 
Thus, $\Delta ABC$ is an equilateral triangle. 
Can you end it now? 
There are number of cases.
A: 
We assume point A and parallel lines L1 and L2 are given.We use the fact that in every triangle  if we draw the diameter of circumscribed circle ,Here from A, the foot of altitude from A and points showing the projection of other vertices on this diameter are cyclic( they are on a circle). So  take two arbitrary points P and Q on lines L1 and L2 respectively.  Draw the circumscribed circle  of triangle APQ and its diameter AD.Draw perpendiculars from P and Q on AD to find E and H which are projections of P and Q on AD. Draw the altitude from A , its foot is K. Draw a circle containing points E, H and K. The center of this circle is coincident on the center of circumscribed circle of triangle APQ. Now if vertex P moves to the left  and Q to the right on lines L1 and L2 in a certain position P, Q and A make an equilateral triangle. As can be seen other two vertices of this triangle B and C are the intersection of the circle passing E, H and K and lines L1 and L2. At this orientation. the projection of C(P), B(Q) and foot of altitude from A( here O) are on a circle with zero diameter which is the center O of circles. Hence a number of these type of triangle can be constructed. 
