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关于△ABC中恒等式ctgA·ctgB+ctgBctgC+ctgCrctgA=1在一般的中学数学书刊中都是通过三角自身的方法去证明的,这里我们用几何的方法给予证明,从而揭示其几何意义。如图1,设△ABC的外心为O,过O点作三边的垂线,分别交三边BC、CA、AB于D、E,F,连OA、OB、OC,因D、E、F分别为三边BC、CA、AB中点,并由圆周角与圆心角关系,则有: ∠A=∠COD。∠B=∠COF
About △ ABC in the identities ctgA · ctgB + ctgBctgC + ctgCrctgA = 1 in the general school mathematics books are to prove through the method of the triangle itself, here we use the geometric method to give proof, thus revealing its geometric significance. As shown in Fig.1, set the outer center of △ABC to O, pass the vertical point of three sides over O point, and pay the three sides BC, CA, AB to D, E, F, and connect OA, OB, OC, because D, E , F are the three sides BC, CA, AB midpoint, and the relationship between the circle angle and the central angle, there are: ∠ A = ∠ COD. ∠B=∠COF