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下述命题是一个常见的几何题,一般的证明方法是本文的证法一;而证法二(见《数学通报》1982。6期,P5,例十,也有其繁杂的一面,对此命题,我想给出一种较简单的证法,也就是本文的证法三。为了比较,对前二种证法略述在前,最后给出证法三。原题由圆外一点A引两切线AS、AT (S、T为切点)。过A引圆的任一割线APQ交ST于R,若M为PQ中点,则AP·AQ=AR·AM. 证法一先证A、S、M、O、T五点共圆,再证△ASR∽△AMS。这是一种常见的证明方法,且证明步骤繁杂。
The following proposition is a common geometric problem. The general proof method is the proof method of this article; and the Proof Code II (see the “Mathematical Bulletin” 1982. 6 period, P5, example 10, also has its complicated side, this proposition I would like to give a simpler proof method, which is also the proof method of this article 3. For comparison, the first two kinds of proofs are outlined and the last one is the proofs of the third. Two tangents AS, AT (S, T are cut-off points), any secant APQ passing over A lead round to ST at R, if M is PQ midpoint, then AP·AQ=AR·AM. The five points A, S, M, O, and T are co-circular, and ΔASR∽ΔAMS is reconfirmed. This is a common method of proof and the procedure is complicated.