The following comments were made by Jim Wilson of the University of Georgia in reply to the student investigation of the problem "Given a quadralateral, construct a point equidistant from its sides."

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Pat --

It is possible to have a quadrilateral ABCD with a point X equidistant from the four sides (meaning equidistant from the lines along which the segments defining the sides will lie) such that X is outside the quadrilateral:

Here is another one with a concave quadrilateral:

Of course, the usual case is to think about a convex quadrilateral. I find it easier to use the incircle and the points of tangency to think about the problem as well as the bisectors of the angles.

Consider A, B, and C given as three of the vertices of a quadrilateral and the angle ABC is determined. Therefore if X is any point on the angle bisector of ABC, a circle can be constructed by dropping a perpendicular to either AB or BC.

Now, construct a point D as the intersection of the lines tangent to the circle from A and from C. Let's name the four tangency points J, K, L, M as in the diagram:

Consider BC - AB. Since AK = AJ, JD = DM, CM = LC, and LB = KB, if follows rather easily that DC -AD = BC - AB.

That is, the difference DC - AD is constant so that A and C are the foci of the hyperbola that is the locus of D.