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平面几何中有关二次方程的问题,大多可以应用韦达定理去解。兹举例如下: 梯形ABCD中(图1),∠B作圆,交BC于E,F。设∠EAB=α,∠EAD=β,求证tgα和tgβ是方程AB·x~2-BC·x+CD=0的两个根。[分析]:在这道题中,只要证明tgα+tgβ=(BC)/(AB),tgαtgβ=(CD)/(AB)就行了。由已知条件,tgα=(BE)/(AB);联DE,∵AD为直径,90°。以AD为直径∠AED=∴tgβ=(DE)/(AE)。但(BE)/(AB)和(DE)/(AE)的分母不同,所以还要化简。联AF,因A、D、F、E四点共圆。∴∠ADE=∠AFE,∠FAB=90°-∠AFE=90°-∠ADE=β,∴tgβ=(BF)/(AB)。因此,解本题的关键在于证
The problem of quadratic equations in plane geometry can mostly be solved by Veda's theorem. Here are some examples: trapezoidal ABCD (Figure 1), 作 B for the circle, cross BC in E, F. Let ∠EAB = α, ∠EAD = β. Verify that tgα and tgβ are two roots of the equation AB · x ~ 2-BC · x + CD = 0. [Analysis]: In this question, tgαtgβ = (CD) / (AB) is ok as long as tgα + tgβ = (BC) / (AB) is proved. From the known conditions, tgα = (BE) / (AB); Associated DE, ∵ AD diameter, 90 °. Take AD as the diameter ∠AED = ∴tgβ = (DE) / (AE). However, the denominators of (BE) / (AB) and (DE) / (AE) are different, so they should be simplified. United AF, because A, D, F, E four points were round. ∴∠ADE = ∠AFE, ∠FAB = 90 ° -∠AFE = 90 ° -∠ADE = β, ∴tgβ = (BF) / (AB). Therefore, the solution to the problem lies in the card