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Discussion The applet may suggest the following statement [Grinberg]:

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

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The proof is by Darij Grinberg based on an idea of Marcello Tarquini.

Let XYZW be the quadrilateral formed be the external angle bisectors of the given quadrilateral ABCD. We know that the perpendiculars to the sides of ABCD through the corresponding vertices of XYZW form an inscriptible quadrilateral. We are going to show that the latter quadrilateral (denoted A'B'C'D' in the applet) is homothetic with A1B1C1D1 wherefrom the proof will follow.

As the first step note that the diagonals of XYZW meet at the incenter of ABCD. This is so because the diagonal XZ is the bisector of the angle formed by BC and AD, and YW is the bisector of the angle between AB and CD. Let M denote the incenter of ABCD.

Since M is the incenter of ABCD, it is the point of concurrency of the internal angle bisectors of ABCD. At any vertex, the internal and external bisectors are perpendicular. Therefore,

 MA ⊥ XW, MB ⊥ XY, MC ⊥ YZ, MD ⊥ ZW.

From here it follows that the quadrilaterals AMBX, BMCY, CMDZ, and DMAW are cyclic. Naturally, their circumcircles meet in M. Furthermore, the first of these has MX as a diameter, with MY, MZ, and MW serving as diameters of the others.

The center Q of the circumcircle of AMBX bisects MX. The perpendicular from Q to the chord AB crosses AB at its midpoint. Therefore, the perpendicular bisector of AB passes through Q, the circumcenter of AMBX. Now, the perpendicular to AB through X is parallel to the perpendicular bisector of AB. One goes through X, the other through Q. Obviously, Q could be obtained from X by the homothety with center M and radius 1/2. Which then implies a less obvious fact that this homothety maps the perpendicular to AB through X onto the perpendicular bisector of AB.

Similar considerations apply to the other sides of ABCD. Thus, the homothety with center M and coefficient 1/2 maps A'B'C'D' onto A1B1C1D1, as promised. Since the former is inscriptible, so is the latter.

This statement has a partial converse. Indeed, if ABCD is cyclic then the perpendicular bisectors of its sides meet in its center - a degenerate quadrilateral which is inscriptible by default, i.e. regardless of whether ABCD is also inscriptible or not. However, if ABCD is not cyclic, and A1B1C1D1 is inscriptible, then so is ABCD.

To prove that, introduce an operator PB - for perpendicular bisectors - such that

 PB(ABCD) = A1B1C1D1

The operator PB has been just shown to have the following property:

 (1) If ABCD is inscriptible, so is PB(ABCD).

 A2B2C2D2 = PB(A1B1C1D1).

We'll show that, for any quadrilateral ABCD, ABCD and PB(PB(ABCD)) are similar. Then, in view of (1), the assumption that PB(ABCD) is inscriptible will imply the same property for PB(PB(ABCD)) and, hence, for ABCD, because of similarity.

Obviously, the sides of A2B2C2D2 are parallel to those of ABCD. But more is true: their diagonals are also parallel. This 6 way parallelism implies that ABCD and A2B2C2D2 are similar. (In fact they are homothetic.) This is because the triangles ABC and A2B2C2 having parallel sides are similar, and the same is true for other pairs of triangles in the two quadrilaterals.

It only remains to show that the diagonals of ABCD and A2B2C2D2 are parallel. A1 is the intersection of the perpendicular bisectors of AB and AD. Therefore, it is the circumcenter of triangle ABD. Such that the perpendicular bisector of BD also passes through A1. Considering the triangle ACD, we similarly have that the same bisector passed through C1. In other words,

 BD ⊥ A1C1,

and similarly,

 AC ⊥ B1D1.

On the second iteration we'll have

 B1D1 ⊥ A2C2 and A1C1 ⊥ B2D2.

Therefore, indeed, AC||A2C2 and BD||B2D2.

### Reference

1. D. Grinberg, A Tour Around Quadrilateral Geometry  