## When A Quadrilateral Is Inscriptible?

For the sake of simplicity, I shall assume throughout that we deal with convex quadrilaterals. If such a quadrilateral ABCD is circumscribed around a circle (a property which is referred to as being inscriptible), then

(1)

AB + CD = BC + AD,

which is easily proven from an observation that two tangents to a circle from a point outside the circle are equal. Thus, (1) is a necessary condition for a (convex) quadrilateral to be inscriptible. Is it sufficient? In other words, given a convex quadrilateral ABCD that satisfies (1), is it inscriptible? Is there a circle that touches all four sides of the quadrilateral? Let us show that this is so:

Any convex quadrilateral that satisfies (1) is inscriptible.

### Proof

We start with a construction of a circle tangent to the sides BC, CD, and AD. (The center of the circle is at the intersection of the bisectors of angles C and D.) If the circle is also tangent to AB we are done. So assume to the contrary that the circle is not tangent to AB. Then find AB' tangent to the circle with B' on BC (or its extension).

Since (1) is a necessary condition for a quadrilateral to be inscriptible and AB'CD does possess this property, we have

(1')

AB' + CD = B'C + AD.

Subtracting (1') from (1) and rearranging the terms we obtain

AB - AB' = BC - B'C = BB',

which contradicts the triangle inequality for ΔABB'.

It ought to be mentioned that the diagram drawn does not cover all possible configurations (e.g., it might have happened that AB crossed the constructed circle), but they all are treated in the same way and lead to a triangle that violates the triangle inequality.

Q.E.D.

This is a classical proof by contradiction or, in Latin, reductio ad absurdum: An assumption that a property holds (or does not hold) leads to a contradiction, therefore the opposite to the assumption must be true.

Looks like we are done and it is time to turn our attention to another interesting topic. However, the proof just given has a flaw that, in principle, may be repaired, but at a considerable effort.

What's wrong with that proof?

The problem with the above proof is in the tacit assumption that the circle touches AD somewhere between A and D. As far as the premises of the theorem go, this fact is not immediately obvious. However, if the point of tangency is outside the segment AD, the whole construction breaks down (especially if it is on the side of A.)

As a replacement, consider the following entirely different approach. However note that a slight conceptual modification of the previous proof in interpreting the symbols in the derivation renders the proof absolutely valid.

First note that any kite is inscriptible, so we may assume that the given quadrilateral ABCD is not a kite. Thus either AD > DC or AD < DC. Assume the former. The latter is considered analogously.

If AD > DC, then (1) implies AB > BC. It is therefore possible to find E on AD and F on AB such that DE = CD and BF = BC.

We can rewrite (1) as

AF + BF + CD = AE + DE + BC.

Taking into account DE = CD and BF = BC we get

AF = AE.

We now see that the three triangles CBF, EDC, and FAE are isosceles. In every isosceles triangle the bisector of the angle at the apex serves as the perpendicular bisector of the base. It follows that the bisectors of the angles at B, A, and D are simultaneously the perpendicular bisectors of the sides of ΔCEF and are, therefore, concurrent. Their common point is equidistant from all four sides of quadrilateral ABCD which leads to the required conclusion that ABCD is inscriptible.

([Andreescu and Enescu, pp. 65-67] gives both proofs without mentioning the flaw in the first one. [Fetisov], on the other hand, makes a point of it.)

### References

1. N. Altshiller-Court, College Geometry, Dover, 1980
2. T. Andreescu, B. Enescu, Mathematical Olympiad Treasures, Birkhäuser, 2004
3. A. I. Fetisov, Ya. S. Dubnov, Proof in Geometry: With "Mistakes in Geometric Proofs", Dover, (December 25) 2006
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