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初中几何第二册P_(85)例1:已知AD是锐角△ABC的高,AE是△AEC外接圆直径,求证:AB·AC=AD·AE。不难证明△ABC为任意三角形时结论亦成立。于是得到一般的结论:三角形任两边之积等于第三边上的高与外接圆直径的乘积。用式于可写成:ab=2R·hc,bc=2R·h_,ca=2R·h_b。在直角三角形中则为ab=chc(c为斜边)。下面举例说明这一公式的应用。例1 已知Rt△ABC中,AB=c,BC=a,AC=b,斜边AB上的高为h,求证:
Junior High School Geometry Booklet P_(85) Example 1: It is known that AD is the height of acute angle △ABC, and AE is the diameter of circumscribed circle of △AEC. Proof: AB·AC=AD·AE. It is not difficult to prove that △ABC is an arbitrary triangle. The general conclusion is thus drawn that the product of any two sides of the triangle is equal to the product of the height of the third side and the diameter of the circumscribed circle. Using the formula can be written as: ab = 2R · hc, bc = 2R · h_, ca = 2R · h_b. In a right triangle, ab=chc (c is a hypotenuse). The following example illustrates the application of this formula. Example 1 It is known that Rt △ ABC, AB = c, BC = a, AC = b, the height of the hypotenuse AB is h, verify: