elem.6.1
同高的三角形和平行四边形，其面积之比等于底边之比。
三角形 ABC、ACD 等高同顶点 A，底 BC、CD 在同一直线上；将 BD 延长至 H、L，依次截取 BG=GH=BC、DK=KL=CD；过 A 平行于 BD 的辅助线上点 E、F 与 BC、CD 构成等高的平行四边形 EC、CF。
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Let ABC, ACD be triangles and EC, CF parallelograms under the same height; I say that, as the base BC is to the base CD, so is the triangle ABC to the triangle ACD, and the parallelogram EC to the parallelogram CF. For let BD be produced in both directions to the points H, L and let [any number of straight lines] BG, GH be made equal to the base BC, and any number of straight lines DK, KL equal to the base CD; let AG, AH, AK, AL be joined. Then, since CB, BG, GH are equal to one another, the triangles ABC, AGB, AHG are also equal to one another.
延长BD至H和L，在BC上取等长线段BG、GH，在CD上取等长线段DK、KL，连接AG、AH、AK、AL。
[I. 38] Therefore, whatever multiple the base HC is of the base BC, that multiple also is the triangle AHC of the triangle ABC. For the same reason, whatever multiple the base LC is of the base CD, that multiple also is the triangle ALC of the triangle ACD; and, if the base HC is equal to the base CL, the triangle AHC is also equal to the triangle ACL, [I. 38] if the base HC is in excess of the base CL, the triangle AHC is also in excess of the triangle ACL, and, if less, less.
由I.38，等底等高的三角形面积相等，故三角形ABC、AGB、AHG面积相等，因此底HC是BC的几倍，三角形AHC就是ABC的几倍。
Thus, there being four magnitudes, two bases BC, CD and two triangles ABC, ACD, equimultiples have been taken of the base BC and the triangle ABC, namely the base HC and the triangle AHC, and of the base CD and the triangle ADC other, chance, equimultiples, namely the base LC and the triangle ALC; and it has been proved that, if the base HC is in excess of the base CL, the triangle AHC is also in excess of the triangle ALC; if equal, equal; and, if less, less. Therefore, as the base BC is to the base CD, so is the triangle ABC to the triangle ACD.
同理，底LC是CD的几倍，三角形ALC就是ACD的几倍；且底HC与CL的大小关系决定三角形AHC与ALC的大小关系。
[V. Def. 5] Next, since the parallelogram EC is double of the triangle ABC, [I. 41] and the parallelogram FC is double of the triangle ACD, while parts have the same ratio as the same multiples of them, [V. 15] therefore, as the triangle ABC is to the triangle ACD, so is the parallelogram EC to the parallelogram FC. Since, then, it was proved that, as the base BC is to CD, so is the triangle ABC to the triangle ACD, and, as the triangle ABC is to the triangle ACD, so is the parallelogram EC to the parallelogram CF, therefore also, as the base BC is to the base CD, so is the parallelogram EC to the parallelogram FC.
由V.定义5得BC:CD = 三角形ABC:三角形ACD；再由I.41及V.15，三角形ABC:ACD = 平行四边形EC:FC，故BC:CD = 平行四边形EC:FC。