Equilateral Triangles On Sides of a Parallelogram II
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Copyright © 1996-2012 Alexander Bogomolny
The applet below provides an illustration to a generalization of a problem that appeared in P. Yiu's article Elegant Geometric Constructions:
Given a rectangle ABCD, to find points P and Q on BC and CD respectively such that ΔAPQ is equilateral, first construct equilateral triangles CDX and BCY, with X and Y inside the rectangle. Then extend AX to intersect BC at P and AY to intersect CD at Q. The triangle APQ is equilateral.
As can be seen from the applet, the construction works for any parallelogram, the main requirement being that triangles BCY and CDX have the same orientation.
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P. Yiu's construction is elegant, but its validation by calculating the lengths of several straight line segments is not. Here is a really elegant substitute.
In the original case where ABCD is a rectangle, ΔAXY is equilateral. (This is because of a property of equilateral triangles constructed on the sides of a parallelogram.) By considering the trapezoid ABCQ, we see that Y is the midpoint of AQ. Similarly, in the trapezoid ADCP, X is the midpoint of AP. Hence,
In general, as professor W. McWorter has observed, ΔAXY is the affine sum of ΔBCY and ΔDXC; it is therefore similar to the two. He also remarked that the similarity of triangles AXY and APQ follows more directly from the proportion
An annonymous visitor has remarked that, in a general case, triangles ABY and XCY are congruent, pretty much by
construction - only
∠AYX = ∠BYC + ∠AYB - ∠CYX = ∠BYC = π/3.
References
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Copyright © 1996-2012 Alexander Bogomolny
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