Equilateral Triangle on Three Lines

Problem

Given three straight lines (denoted below by two points $AB,$ $CD,$ $EF$). Construct an equilateral triangle with vertices one per line.

Construction

Choose an arbitrary point on one of the line, say $X$ on $EF.$ Rotate $CD$ around $X$ $60^{\circ}$ into $C'D'.$ Let $Y'$ be the intersection of $C'D'$ with $AB$ and $Y$ the point that was mapped into $Y'$ by the rotation. Then $\Delta XYY'$ is equilateral and $X\in EF,$ $Y'\in AB,$ and $Y\in CD.$

In general, the solution is not unique, e.g., Acknowledgment

The construction is classic. For three parallel lines (when the solution is unique up to isomorphism), it can be found in I. M. Yaglom, Geometric Transformations I (MAA, 1962), problem 18. The latter problem has been discussed elsewhere.

Various Geometric Constructions

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• The Many Ways to Construct a Triangle and additional triangle facts
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• Construction of n-gon from the midpoints of its sides
• Short Construction of the Geometric Mean
• Construction of a Polygon from Rotations and their Centers
• Squares Inscribed In a Triangle I
• Construction of a Cyclic Quadrilateral
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• Tangent to Circle in Three Steps
• Regular Pentagon Construction by K. Knop
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