Perpendicular Bisectors in an Inscriptible Quadrilateral: What is this about?
A Mathematical Droodle

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

|Activities| |Contact| |Front page| |Contents| |Geometry|

Copyright © 1996-2018 Alexander Bogomolny

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.

(A zipped Sketchpad sketch is available for download.)

References

  1. M. de Villiers, Private communication, Dec. 2004
  2. M. de Villiers, Some Adventures in Euclidean Geometry, Univ. of Durban-Westville, 1994 (revised 1996), pp. 192-193

|Activities| |Contact| |Front page| |Contents| |Geometry|

Copyright © 1996-2018 Alexander Bogomolny

72386318