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Given a triangle ABC, a new triangle A'B'C' can be formed as follows: Drawthe perpendicular to the line AB at the pointA; then the perpendicular to theline BC at B, and lastly the perpendicular to the line CA at C.the twotriangles ABC and A'B'C' are always similar. In Postulate1 we prove that theratio E'/E is equal to (cotA+cotB+cotC)^2, which is the main result in thiswork.Here E' and E stand for the areas of the triangles A'B'C' and ABCrespectively. 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'), withboth ABC and A'B'C' being right triangles, then the minimum value of the aboveratio of areas, is 4.

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