  Given a triangle ABC, a new triangle A'B'C' can be formed as follows: Draw
the perpendicular to the line AB at the pointA; then the perpendicular to the
line BC at B, and lastly the perpendicular to the line CA at C.the two
triangles ABC and A'B'C' are always similar. In Postulate1 we prove that the
ratio E'/E is equal to (cotA+cotB+cotC)^2, which is the main result in this
work.Here E' and E stand for the areas of the triangles A'B'C' and ABC
respectively. In Postulate 2, we show that the above ratio has minimum value 3,
which is attained when ABC and A'B'C' are equilateral triangles.In Postulate 3,
we show that if we consider only those pairs of triangles (ABC,A'B'C'), with
both ABC and A'B'C' being right triangles, then the minimum value of the above
ratio of areas, is 4.
