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中学几何中,有一类关于证明1/a+1/b=t/c型线段关系式的题目,这类问题的解决,一般地归结为寻求c/a和c/b的具有公分母的“第三比例式”,这里再介绍两种方法。 一、将1/a+1/b=t/c型化为e/f=h/g型解决(其中a,b,c,t,e,f,h,g都是线段) 例1.BF是△ABC中∠ABC的平分线,过F作EF∥BC交AB于E,则1/AB+1/BC=1/BE。
In the middle school geometry, there is a class of questions that certify the relationship of the 1/a+1/b=t/c type line segment. The solution to this type of problem generally comes down to the common denominator that seeks c/a and c/b. The third ratio ", here we introduce two methods. 1. Resolve 1/a+1/b=t/c to e/f=h/g (where a,b,c,t,e,f,h,g are line segments) Example 1. BF is the bisector of ABC in △ABC. If F is used as EF, BC is transferred to AB in E, then 1/AB+1/BC=1/BE.