灯下 登录
数学 / 几何原本 / Proposition VI.27

第6卷命题 27 · 最大平行四边形定理

Of all the parallelograms applied to the same straight line and deficient by parallelogrammic figures similar and similarly situated to that described on the half of the straight line, that parallelogram is greatest which is applied to the half of the straight line and is similar to the defect.

在同一直线上所有内接平行四边形中,若其缺损部分与半线段上的平行四边形相似且相似放置,则半线段上的平行四边形最大。

A B C D E F G H K L M N
fig-1

在 AB 上以与 AB 一半上类似平行四边形为亏形作贴合:贴在中点 D 半边 AD 上的图形 EF 面积最大,K、L、M、N 等其他位置的贴合图形面积均小于它。

线

正文图形由校订坐标生成;点、线、角、圆可与证明和问答联动。

分步证明Step-by-step proof
1 / 4
  1. Let AB be a straight line and let it be bisected at C; let there be applied to the straight line AB the parallelogram AD deficient by the parallelogrammic figure DB described on the half of AB, that is, CB; I say that, of all the parallelograms applied to AB and deficient by parallelogrammic figures similar and similarly situated to DB, AD is greatest. For let there be applied to the straight line AB the parallelogram AF deficient by the parallelogrammic figure FB similar and similarly situated to DB; I say that AD is greater than AF.

    设AB被C平分,AD是应用在AB上的平行四边形,其缺损DB与CB上的平行四边形相似。

  2. For, since the parallelogram DB is similar to the parallelogram FB, they are about the same diameter. [VI. 26] Let their diameter DB be drawn, and let the figure be described.

    任取另一平行四边形AF,其缺损FB与DB相似且相似放置。

  3. Then, since CF is equal to FE, [I. 43] and FB is common, therefore the whole CH is equal to the whole KE. But CH is equal to CG, since AC is also equal to CB.

    由于DB与FB相似,它们有相同的对角线,设对角线为DB,并完成图形。

  4. [I. 36] Therefore GC is also equal to EK. Let CF be added to each; therefore the whole AF is equal to the gnomon LMN; so that the parallelogram DB, that is, AD, is greater than the parallelogram AF.

    由面积相等关系可得AD大于AF,因此AD是最大的。