Carnot's Theorem
Carnot's Theorem states that in a triangle with
,
, and
, perpendiculars to the sides
,
, and
at
,
, and
are concurrent if and only if
.
Contents
[hide]Proof
Only if: Assume that the given perpendiculars are concurrent at . Then, from the Pythagorean Theorem,
,
,
,
,
, and
. Substituting each and every one of these in and simplifying gives the desired result.
If: Consider the intersection of the perpendiculars from and
. Call this intersection point
, and let
be the perpendicular from
to
. From the other direction of the desired result, we have that
. We also have that
, which implies that
. This is a difference of squares, which we can easily factor into
. Note that
, so we have that
. This implies that
, which gives the desired result.
Problems
Olympiad
is a triangle. Take points
on the perpendicular bisectors of
respectively. Show that the lines through
perpendicular to
respectively are concurrent. (Source)