## 9889Re: [EMHL] The Brocard points of a quadrilateral (Geoff Millin)

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• Jun 9, 2004
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
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