Joseph Keech in Bride's Chair

A former correspondent of mine has recently cc'ed me a copy of an old (1783) article with a statement of properties of the famous Bride's Chair configuration, with the top (or bottom, depending on the outlook) part missing:

Let ABC be a plane triangle, right angled at C; and let the squares ACDE and BCKL be described on the two legs AC and BC; also let the straight lines AL and BE be drawn from the two acute angles to the opposite angles L and E of the two squares, cutting the legs of the triangle in F and H; I say that CF shall be equal to CH and each of them to the side of a square HCFZ, inscribed in the triangle ABC.

 

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Solution

References

  1. Joseph Keech, in Gentleman's Monthly Intelligencer, July, 1783, 11-13

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Since AC = CD and BC = CK, AK = BD. And because the triangles AKL, ACF are similar, as well as triangles BDE and BCH, AK/AC = KL/CF and BD/BC = DE/CH. Now, as the three first terms in each proportion are respectively equal, the last must be equal also, that is CF = CH.

 

This applet requires Sun's Java VM 2 which your browser may perceive as a popup. Which it is not. If you want to see the applet work, visit Sun's website at https://www.java.com/en/download/index.jsp, download and install Java VM and enjoy the applet.


What if applet does not run?

Draw FZ (Z on AB) parallel to AC, and, consequently, to BL, also join HZ. Then because the triangles ABL and AZF are similar, (KL/CF =) AL/AF = BL/FZ. Hence, as KL = BL, CF = FZ = CH; consequently HZ is equal and parallel to CF, and the figure HCFZ is equilateral. Moreover, the angles at C and F being right angles by construction, the opposite ones at Z and H are right angles also and HCFZ is a square.

J. Keech's note proves another property of the configuration (check in the applet the "extra" box):

If the same construction remain, and if the figure HCFZ be circumscribed by the circle HCFZ, meeting AB of the triangle again in G; and if GC, GF, and GH be drawn: I say that the angles FGB, FGC, HGC, and HGA are each of them equal to half a right angle.

I came across this result some time ago in a modern book of geometry problems and have treated it elsewhere.


Related material
Read more...

  • Bride's Chair
  • Bride's Chair: an Interactive Gizmo
  • Desargues in the Bride's Chair(with Pythagoras)
  • Concyclic Points in Bride's Chair
  • A Square in the Chair
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