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中考数学试卷中a·b=c·d±e·f型平面几何题是一类分值多、难度大的问题.本文将用因式分解这一代数变换,探寻其证题思路.思想方法:要证a·b=c·d+e·f,可将其中两项组合并分解因式,使结论转化为只含有四条线段的等积式或两条线段相等.例1:已知,如图1,RtΔABC中,∠A=90°,D是AB上一点,以BD为直径的圆交BC于E,连CD并延长交圆于F.求证:BD~2=BC·BE-CD·DF(1993年遵义)分析:连结DE、FB,∵BD是直径,∴∠CED=∠A=∠F=90°,∴A、C、E、D共圆,∴BE·BC=BD·BA,原结论转化为:BD~2=BD·BA-CD·DF移项后分解因式得:BD(BA-BD)=CD·DF即BD·
The a · b = c · d ± e · f plane geometry questions in the test paper of the senior high school entrance examination are a kind of problems with a great deal of points and great difficulty.This paper will use the algebraic transformation of factorization to explore the way of thinking of the syndromes. : To prove a · b = c · d + e · f, two of them can be combined and factorized, so that the conclusion is converted into an equivariant or two segments equal to only four segments.Example 1: It is known that, As shown in Figure 1, RtΔABC, ∠A = 90 °, D is AB on the previous point, with BD as the diameter of the circle BC in E, with CD and extend the circle at F. Verify: BD ~ 2 = BC · BE-CD DF (1993 Zunyi) Analysis: Link DE, FB, ∵BD is the diameter, ∴∠CED = ∠A = ∠F = 90 °, ∴A, C, E, D common circle, ∴BE · BC = BD · BA, the original conclusion is transformed into: BD ~ 2 = BD · BA-CD · DF After the shift factorization formula was: BD (BA-BD) = CD · DF that BD