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数学 / 几何原本 / Proposition I.40

第1卷命题 40 · 等底同侧相等三角形在同一平行线间

Equal triangles which are on equal bases and on the same side are also in the same parallels.

等底同侧的相等三角形,其顶点在同一条与底边平行的直线上。

A B C D E F
fig-1

等积三角形 ABC、DEF 在相等底 BC、EF 上同侧;要证 A、D 两顶点在同一条与底平行的直线上。

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
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  1. Let ABC, CDE be equal triangles on equal bases BC, CE and on the same side. I say that they are also in the same parallels.

    把等底相等三角形放在同侧。

  2. For let AD be joined; I say that AD is parallel to BE. For, if not, let AF be drawn through A parallel to BE [I. 31], and let FE be joined.

    若顶点不在同一条与底平行的直线上,则用等底构造相同平行线间的比较三角形。

  3. Therefore the triangle ABC is equal to the triangle FCE; for they are on equal bases BC, CE and in the same parallels BE, AF. [I. 38] But the triangle ABC is equal to the triangle DCE; therefore the triangle DCE is also equal to the triangle FCE, [C.N. 1] the greater to the less: which is impossible.

    euclid-elements/book1-prop-038 会推出面积不等。

  4. Therefore AF is not parallel to BE. Similarly we can prove that neither is any other straight line except AD; therefore AD is parallel to BE.

    矛盾说明两个顶点必须在同一条平行线上。