# Equilateral Triangle on Three Lines

17 July 2014, Created with GeoGebra

### 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

• How to Construct Tangents from a Point to a Circle
• How to Construct a Radical Axis
• Constructions Related To An Inaccessible Point
• Inscribing a regular pentagon in a circle - and proving it
• The Many Ways to Construct a Triangle and additional triangle facts
• Easy Construction of Bicentric Quadrilateral
• Easy Construction of Bicentric Quadrilateral II
• Star Construction of Shapes of Constant Width
• Four Construction Problems
• Geometric Construction with the Compass Alone
• 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
• Circle of Apollonius
• Six Circles with Concurrent Pairwise Radical Axes
• Trisect Segment: 2 Circles, 4 Lines
• Tangent to Circle in Three Steps
• Regular Pentagon Construction by K. Knop