### 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?

Discussion The applet may suggest the following statement [de Villiers]:

 Given a qudrilateral ABCD, the perpendicular bisectors of its sidelines form a quadrilateral A1B1C1D1. If ABCD is inscriptible, then so is A1B1C1D1.

### 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?

The applet is suggestive of a statement but does not seem to help with finding a proof. The theorem could be derived from a similar theorem concerned with angles bisectors. A presentation of the proof with a different illustration appears elsewhere. Below, I give a proof kindly supplied to me by M. de Villiers. The proof is included in his book Some Adventures in Euclidean Geometry and was suggested by Jordan Tabov.

I apologize for a change of notations. As seen in the diagram, the given quadrilateral ABCD is inscriptible. X, Y, Z, W are the midpoints of the sides AD, AB, BC, and CD, respectively. HY ⊥ AB, HZ ⊥ BC, FX ⊥ AD, and FW ⊥ CD. E is the point of intersection HY and FX, G is the intersection of FW and HZ, and M is the intersection of FX nc CD.

We want to show that the quadrilateral EFGH is inscriptible. Let AB = a, BC = b, CD = c, and DA = d. tA is the tangent from A to the incircle. tB, tC, tD are defined similarly. Assume the radius of the incircle equals 1.

We now have:

 (1) 2·MD = -d/cosD, 2·MW = c - d/cosD, 2·FW = 2·MW·cot(180° - D) = d/sinD - c·cotD.

Similarly

 (2) 2·FX = c/sinD - d·cotD.

Hence

(3)
 2(FX - FW) = (c-d)/sinD + (c-d)cotD = (c-d)(1 + cosD)/sinD = (c-d)cot(D/2) = (c-d)tD/r = (c-d)tD.

Simialrly for 2(HZ - HY), 2(EY - EX) and 2(GW - GZ). We now have to check that the sums of the opposite sides of EFGH are equal, or alternatively that

 EF - EH + HG - GF = 0,

which is the same as

 (FX-EX) - (HY-EY) + (HZ-GZ) - (FW-GW) = 0.

Rearranging the terms and multiplying by 2 gives

 (4) 2(FX - FW) + 2(HZ - HY) + 2(EY - EX) + 2(GW - GZ) = 0.

In view of (3), (4) is equivalent to

 (5) (c - d)tD + (d - a)tA + (a - b)tB + (b - c)tC = 0.

Since ABCD is inscriptible, c - d = b - a and, equivalently, d - a = c - b. With this in mind equation (5) becomes

 (6) (c - d)(tD - tB) + (d - a)(tA - tC) = 0.

But

 c - d = (tC + tD) - (tD) + tA) = tC - tA.

and similarly

 d - a = tD - tB.

Therefore (6) is equivalent to

 (tC - tA)(tD - tB) + (tD - tB)(tA - tC) = 0.

which is an identity and completes the proof. 