Area of a circle as follows. Draw a perpendicular from the center to the midpoint of a side of the polygon; its length, "h", is less than the circle radius. Also, let each side of the polygon have length "s"; then the sum of the sides, "ns", is less than the circle circumference. The polygon area consists of "n" equal triangles with height "h" and base "s", thus equals ⁄"nhs". But since "h" < "r" and "ns" < "c", the polygon area must be less than the triangle area, ⁄"cr", a contradiction. Therefore, our supposition that "C" might be greater than "T" must be
Distance from a point to a line a side, and "h" is the perpendicular height from the opposite vertex. This proof is only valid if the line is neither vertical nor horizontal, that is, we assume that neither "a" nor "b" in the equation of the line is zero. The line with equation "ax" + "by" + "c" = 0 has slope -"a"/"b", so any line perpendicular to it will have slope "b"/"a" (the negative reciprocal). Let ("m", "n") be the point of intersection of the line "ax" + "by" + "c" = 0 and the line perpendicular to it which passes through the point ("x", "y").