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be a convex quadrilateral. The perpendicular bisectors of its sides 
and 
meet at 
. Denote by 
a point inside the quadrilateral such that 
and 
. Show that 
.
such that
for all nonnegative real numbers 
, 
, and 
with 
.
be any point in the plane of triangle 
. Let 
be the inradius of , let 
, 
, 
be the lengths of the altitudes from 
, 
, 
, respectively, and let 
for 
. Prove 
.
is cyclic and 
, we have ![\[\angle MAC=\angle MAQ=\angle MPQ=90^{\circ}-\angle APQ=90^{\circ}-\angle CBA.\]](http://latex.artofproblemsolving.com/9/1/2/9120de1ae5672cd215e27c8ceedf7e64dd25fccc.png)
be the second intersection of the median 
with the circumcircle of . Then ![\[ \angle HBA=\angle HBC+\angle CBA=\angle HAC+\angle CBA=90^{\circ},\]](http://latex.artofproblemsolving.com/0/2/c/02c11285af50f2280d86b29773322ba19b230a1d.png)
so 
is a diameter of the circumcircle. Let 
be the center of this circumcircle. Then we know 
are colinear. If 
, then 
, so 
and is isosceles, but we assumed it was scalene. Thus 
, and so 
contains , so 
is a diameter of the circumcircle of , so 
.
