# 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

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.

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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.