Line Work - Futility Closet

bilinski problem

Robert Bilinski proposed this problem in the April 2006 issue of Crux Mathematicorum. On square ABCD, two equilateral triangles are constructed, ABE internally and BCF externally, as shown. Prove that D, E, and F are collinear.

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This solution is by Titu Zvonaru. Triangle ADE is isosceles (because AD and AE are the same length). And ∠DAE = 90° – 60° = 30°. Similarly, triangle EBF is isosceles (because BE and BF are the same length). And ∠EBF = 60° + 30° = 90°. So $\displaystyle \angle \mathit{DEF} = \angle \mathit{DEA} + \angle \mathit{AEB} + \angle \mathit{BEF} \\\\ = \frac{180^{\circ} - 30^{\circ}}{2} + 60^{\circ} + \frac{180^{\circ} - 90^{\circ}}{2} = 180^{\circ}.$ Hence D, E, and F are collinear. 04/03/2022 UPDATE: Reader Peter Smith sent this alternative proof, which doesn’t require the calculation of angles: AECFB is a symmetrical pentagon (containing two equilateral triangles); therefore EC is parallel to AF. It follows that ∠DCE is equal to ∠BAF. By symmetry about a vertical axis, ∠CDE is equal to ∠DCE. By symmetry about a horizontal axis, ∠CDF is equal to ∠BAF. Therefore ∠CDE is equal to ∠CDF and DEF must be collinear. (Thanks, Peter.)

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