theorem
  a^2 + b^2 = 1 & c^2 + d^2 = 1 & c * a + d * b = 1 implies b * c = a * d
  proof
    assume that
A1: a^2 + b^2 = 1 and
A2: c^2 + d^2 = 1 and
A3: c * a + d * b = 1;
    c^2 + d^2 = (c * a + d * b)^2 by A3,A2
             .= c^2 * a^2 + 2 * (c * a) * (d * b) + d^2 * b^2;
    then (1 - a^2) * c^2 + (1 - b^2) * d^2 - 2 * (c * a) * (d * b) = 0;
    then (b * c)^2 + a^2 * d^2 - 2 * (c * a) * (d * b) = 0 by A1,SQUARE_1:9;
    then ((b * c) - (a * d))^2 = 0;
    then (b * c) - (a * d) = 0;
    hence thesis;
  end;
