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数学 / 几何原本 / Proposition V.24

第5卷命题 24 · 第五卷第二十四命题

elem.5.24

若第一量比第二量等于第三量比第四量,且第五量比第二量等于第六量比第四量,则第一量与第五量之和比第二量等于第三量与第六量之和比第四量。

A B C D E F G H
fig-1

若 A:C = B:D 且 E:C = F:D,则 (A+E):C = (B+F):D(同后项相加)。G、H 为辅助等倍量。

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分步证明Step-by-step proof
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  1. Let a first magnitude AB have to a second C the same ratio as a third DE has to a fourth F; and let also a fifth BG have to the second C the same ratio as a sixth EH has to the fourth F; I say that the first and fifth added together, AG, will have to the second C the same ratio as the third and sixth, DH, has to the fourth F. For since, as BG is to C, so is EH to F, inversely, as C is to BG, so is F to EH.

    设第一量AB比第二量C等于第三量DE比第四量F,且第五量BG比第二量C等于第六量EH比第四量F。

  2. Since, then, as AB is to C, so is DE to F, and, as C is to BG, so is F to EH, therefore, ex aequali, as AB is to BG, so is DE to EH.

    由BG:C = EH:F,反比得C:BG = F:EH。

  3. [V. 22] And, since the magnitudes are proportional separando, they will also be proportional componendo; [V. 18] therefore, as AG is to GB, so is DH to HE.

    由AB:C = DE:F及C:BG = F:EH,根据等比定理(V.22)得AB:BG = DE:EH。

  4. But also, as BG is to C, so is EH to F; therefore, ex aequali, as AG is to C, so is DH to F.

    由AB:BG = DE:EH,根据合比定理(V.18)得AG:GB = DH:HE;又BG:C = EH:F,再次应用等比定理得AG:C = DH:F。