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1 命题及其证明命题 如图 1所示 ,若直线 l⊥线段 AB于 H ,则M1 A2 - MA2 =M1 B2 - MB2 (1)反之 ,若式 (1)成立 ,则 M1 M所在的直线 l⊥AB.图 1证明 ∵ l⊥线段AB,∴ 由勾股定理得 :AM21 - AH 2 =H M21 ,AM2 - AH 2 =H M2 .两式相减得AM21 - AM2 =H M21 - H M2 . 1同理可
1 The proposition and its proposition are shown in Fig. 1. If the straight line l⊥ AB is in H, M1 A2 - MA2 = M1 B2 - MB2 (1) Conversely, if the formula (1) holds, the line where M1 M is located ⊥ AB. Figure 1 proves that the line segment AB of ∵ l ∴ is obtained from the Pythagorean Theorem: AM21 - AH2 = H M21, and AM2 - AH2 = H M2. The two equations subtract AM21 - AM2 = H M21 - H M2. 1 same reason can