>

>Given a cyclic quadrilateral ABCD, let S be the meet

>of the diagonals AC and BD, and K and L the

>orthogonal projections of S on the sides AB and CD.

>Show that the perpendicular bisector of the segment

>KL bisects the sides BC and DA.

>

Dear Darij, Jean-Pierre and other colleagues!

There is another interesting problem. Let given a quadrilateral ABCD, S is

the common point of AC and BD, K, L, M, N - the projections of S on AB, BC,

CD, DA. Then if we have only the points K, L, M, N we can restore the

quadrilateral ABCD. So we have a bijection of quadrilaterals ABCD - KLMN.

The problem is: what properties of KLMN corresponds to given properties of

ABCD. Some results are known.

1. KLMN is circumscribed if and only if ABCD is inscribed.

2. KLMN is inscribed if and only if AC and BD are perpendicular.

3. KL*MN=ML*NK if and only if PS and SQ are perpendicular, where P is common

point of AB and CD, Q - common point of AD and BC.

4. KL and MN are parallel if and only if the angles A and C are equal.

5. The angles K and M are equal if and only if AD and BC are parallel.

Does anybody know another facts?

My paper concerning this problem was typed in "Kvant" (1998, N 4).

Sincerely Alexey