Carnot's perpendicularity theorem gives a criterion for three lines, each perpendicular to an extended side
of a triangle, to be concurrent.
Let 
,
,
and 
lie on the lines 
, 
, and 
, respectively. The perpendiculars
at 
,
,
and 
are concurrent iff
 |
where the lengths are ordinary unsigned distances (Bogomolny).
If the perpendiculars meet at 
, the Pythagorean theorem
gives
 |
Adding this relation and its cyclic analogs proves the criterion. Conversely, let 
be the intersection of the perpendiculars
at 
and 
,
and let 
be its perpendicular foot
on 
.
The criterion for 
, 
, 
then gives 
. On the fixed line 
,
this difference is a nonconstant linear function
of position, so 
and the third perpendicular
also passes through 
.
Taking 
,
,
and 
to be the midpoints of the sides
makes all three differences vanish. Thus the perpendicular
bisectors are concurrent at the circumcenter.
