A Square in the Chair: What is this about?
A Mathematical Droodle

The applet below has just a few controls. The shape of a right triangle can be changed by dragging its vertices. More importantly, the whole triangle can also be dragged to another location within the applet area.

Explanation

The applet is assumed to hint at a property of the Bride's Chair observed by Douglas Rogers. The observation only needs the squares on the legs of a right triangle. The square on the hypotenuse serves as a reminder of the story's background. (The notations in the applet and below are the same as in the Proof #1 of the Pythagorean theorem.

First of all, recollect that the two lines AK and BF meet on the altitude of the triangle dropped from angle C. This is true for any Vecten's configuration even where the angle at C is not right. Let AK and BK cross BC and AC in S and R, respectively. It is known that CR = CS. In fact, more is true: the vertex M of the implied square CRMS lies on the hypotenuse AB. (So this is exactly the square that made its appearance in Proof #76 of the Pythagorean theorem.

Let's see why this is so. Assume first that M' is the intersection with AB of the perpendicular to BC at S. ΔACS is similar to both ΔKBS and ΔM'SB. This leads to the following proportions:

Combining the two gives BS/CS = BS/M'S which implies CS = M'S. Now, with x = CS, the first of the proportions in (1) can be rewritten as

which is the same as 1 - x/a = x/b implying

Assume now M'' is the intersection with AB of the perpendicular to AC at R. Similarly to the above we obtain CR = M''R = ab/(a + b). Forming a square CRMS we see that M lies on both perpendiculars M'S and M''R and, in addition,

which shows that all three points M, M', and M'' coincide and so M does lie on AB.

The statement just proved shows that the points of intersection of the sides AC and BC with the rafters AK and BF of the Bride's Chair serve as two vertices of a square CRMS inscribed into ΔABC. The last vertex M is naturally located at the foot of the bisector of the right angle.

Prof. W. McWorter has suggested a different approach. Let M° be the foot of the bisector of the right angle at C. Drop the perpendiculars M°R° and M°S° on sides AC and BC, respectively. Let X denote the remaining, unmarked vertex of the square on side AC. Then two trapezoids ABXF and M°BCR° are similar and similarly oriented. It follows that the angles of the trapezoids at B are equal and, since sides BC and BX coincide, so are the two lines BF and BR°. (This in fact means that the two trapezoids are homothetic from B.) The same applies to the other pair trapezoids (homothetic from A) giving that S° lies on BC.

References

1. J. Casey, A Sequel to Euclid, Scholarly Publishing Office, University of Michigan Library (December 20, 2005), reprint of the 1888 edition, p. 17.