In Hyacinthos message #9843, I wrote:

>> This is very interesting! At first, consider

>> the Brocard points R and R' of the harmonic

>> quadrilateral ABCD, and the symmedian point

>> K (this is the point whose distances to the

>> sides of the quadrilateral are proportional

>> to the lengths of the sides). Also consider

>> the circumcenter O of our quadrilateral.

>> Then, the points O, K, R and R' lie on one

>> circle, which is analogous to the "Brocard

>> circle" of a triangle. The points O and K

>> are the endpoints of a diameter of this

>> circle. The points R and R' lie

>> symmetrically with respect to this diameter.

>> The circle through the points O, K, R and R'

>> also passes through the midpoints of the

>> diagonals AC and BD. The line joining these

>> midpoints meets the diameter OK at the

>> midpoint of the segment RR'. The angles

>> < ROK and < R'OK are equal to the Brocard

>> angle of the quadrilateral ABCD.

These remarks started an interesting thread,

which I could not follow since one half of the

mathematics involved was new to me, but let me

just make one remark: In a harmonic

quadrilateral ABCD, the point K is the point

of intersection of the diagonals AC and BD.

[This is quite easy to prove.] Now, since the

midpoints M and N of the diagonals AC and BD

are the orthogonal projections of the

circumcenter O on these diagonals, the angles

< OMK and < ONK are right angles. Hence, M and

N lie on the circle with diameter OK.

Darij Grinberg