On an area property of the sum cotA+cotB+cotC in a triangle
  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. 
Konstantine Zelator