第6卷命题 22 · 比例线段上相似形比例定理

Let the four straight lines AB, CD, EF, GH be proportional, so that, as AB is to CD, so is EF to GH, and let there be described on AB, CD the similar and similarly situated rectilineal figures KAB, LCD, and on EF, GH the similar and similarly situated rectilineal figures MF, NH; I say that, as KAB is to LCD, so is MF to NH. For let there be taken a third proportional O to AB, CD, and a third proportional P to EF, GH. [VI. 11] Then since, as AB is to CD, so is EF to GH, and, as CD is to O, so is GH to P, therefore, ex aequali, as AB is to O, so is EF to P.

设AB、CD、EF、GH成比例，即AB:CD=EF:GH。在AB、CD上作相似且相似放置的直线形KAB、LCD，在EF、GH上作相似且相似放置的直线形MF、NH。

[V. 22] But, as AB is to O, so is KAB to LCD, [VI. 19, Por.] and, as EF is to P, so is MF to NH; therefore also, as KAB is to LCD, so is MF to NH. [V. 11] Next, let MF be to NH as KAB is to LCD; I say also that, as AB is to CD, so is EF to GH. For, if EF is not to GH as AB to CD, let EF be to QR as AB to CD, [VI. 12] and on QR let the rectilineal figure SR be described similar and similarly situated to either of the two MF, NH.

取AB、CD的第三比例项O，以及EF、GH的第三比例项P。由比例性质，AB:O=EF:P。

[VI. 18] Since then, as AB is to CD, so is EF to QR, and there have been described on AB, CD the similar and similarly situated figures KAB, LCD, and on EF, QR the similar and similarly situated figures MF, SR, therefore, as KAB is to LCD, so is MF to SR. But also, by hypothesis, as KAB is to LCD, so is MF to NH; therefore also, as MF is to SR, so is MF to NH. [V. 11] Therefore MF has the same ratio to each of the figures NH, SR; therefore NH is equal to SR.

根据相似形面积比等于对应边平方比，有KAB:LCD=AB:O，MF:NH=EF:P，因此KAB:LCD=MF:NH。

[V. 9] But it is also similar and similarly situated to it; therefore GH is equal to QR. And, since, as AB is to CD, so is EF to QR, while QR is equal to GH, therefore, as AB is to CD, so is EF to GH.

反之，若KAB:LCD=MF:NH，假设EF:GH≠AB:CD，作EF:QR=AB:CD，并在QR上作与MF、NH相似的SR。则KAB:LCD=MF:SR，结合假设得MF:SR=MF:NH，故SR=NH，从而QR=GH，推出AB:CD=EF:GH。