Any two intersecting chords of a circle will each get divided into two segments. If one chord has segments of length A and B, and the other chord has segments of length C and D, then A*B = C*D because of the similar triangles that are formed.
Thus, if the diagonals of a quadrilateral have the same property, it must be a cyclic quadrilateral because the diagonals are chords of a circle touching the corners of the quadrilateral.
However, except for a square, the intersection of the diagonals or chords cannot be the center of the circle because that would make two pairs of identical triangles.
Lockhart noted that if one connects the midpoints of the sides of ANY quadrilateral, the result is a parallelogram.
What Lockhart ought to investigate is whether the diagonals of any quadrilateral are chords of a circle. Of course, the answer is no, but why?
The proof is left as an exercise for the reader.
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