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韦达定理不仅在代数中有广泛的应用,还可用于证明平面几何题,且看下例: P是等边三角形ABC外接圆的劣弧■上任一点(包括极限位置B、C)求证: ① PB+PC=PA; ② PB·PC=(PA~2-AB~2)。分析由结论知,PB、PC可视作某一元二次方程的二根,由韦达定理,知此方程是 x~2-PA·x+(PA~2-AB~2)=0。且PB、PC满足此方程,即得ⅰ) PB~2-PA·PB+(PA~2-AB~2)=0, ⅱ) PC~2-PA·PC+(PA~2-AB~2)=0。如能证明以上二式,命题即可得证。如图1,在△ABP中,依余弦定理得 AB~2=PA~2+PB~2-2PA·PBcos60°,
The Weddah theorem is not only widely used in algebra, but it can also be used to prove plane geometry problems. Let’s look at the following example: P is the inferior arc of the circumcircle of the equilateral triangle ABC. Verify at any point (including the extreme positions B and C): 1 PB+PC=PA; 2 PB·PC=(PA~2-AB~2). The analysis is known by the conclusion that PB and PC can be regarded as the two quadratic equations of a certain element. From the Weida theorem, it is known that this equation is x~2-PA·x+(PA~2-AB~2)=0. PB and PC satisfy this equation, ie, i) PB~2-PA·PB+(PA~2-AB~2)=0, ii) PC~2-PA·PC+(PA~2-AB~2)= 0. If you can prove the above two types, the proposition can be obtained. As shown in Figure 1, in △ABP, according to cosine theorem AB~2=PA~2+PB~2-2PA•PBcos60°,