To a given straight line to apply a parallelogram equal to a given rectilineal figure and exceeding by a parallelogrammic figure similar to a given one.
在给定线段上作一个平行四边形，使其等于已知直线形，且超出部分是一个与给定平行四边形相似的平行四边形。
在直线 AB 上贴合与图形 C 等积、超出部分与已知平行四边形 D 相似的平行四边形：辅助点 E～P 与下方 V、W、X 完成与命题 28 对偶的构造。
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Let AB be the given straight line, C the given rectilineal figure to which the figure to be applied to AB is required to be equal, and D that to which the excess is required to be similar; thus it is required to apply to the straight line AB a parallelogram equal to the rectilineal figure C and exceeding by a parallelogrammic figure similar to D. Let AB be bisected at E; let there be described on EB the parallelogram BF similar and similarly situated to D; and let GH be constructed at once equal to the sum of BF, C and similar and similarly situated to D. [VI. 25] Let KH correspond to FL and KG to FE.
设AB为给定线段，C为已知直线形，D为给定平行四边形。将AB平分于E，在EB上作平行四边形BF相似于D且位置相同。
Now, since GH is greater than FB, therefore KH is also greater than FL, and KG than FE. Let FL, FE be produced, let FLM be equal to KH, and FEN to KG, and let MN be completed; therefore MN is both equal and similar to GH. But GH is similar to EL; therefore MN is also similar to EL; [VI. 21] therefore EL is about the same diameter with MN.
作平行四边形GH等于BF与C之和，且相似于D。使KH对应FL，KG对应FE。
[VI. 26] Let their diameter FO be drawn, and let the figure be described. Since GH is equal to EL, C, while GH is equal to MN, therefore MN is also equal to EL, C. Let EL be subtracted from each; therefore the remainder, the gnomon XWV, is equal to C.
延长FL、FE，使FLM等于KH，FEN等于KG，完成平行四边形MN。则MN等于且相似于GH，而GH相似于EL，故MN相似于EL，因此EL与MN同对角线。
Now, since AE is equal to EB, AN is also equal to NB [I. 36], that is, to LP [I. 43]. Let EO be added to each; therefore the whole AO is equal to the gnomon VWX. But the gnomon VWX is equal to C; therefore AO is also equal to C.
作对角线FO，完成图形。由于GH等于EL加C，且GH等于MN，故MN等于EL加C。减去EL，剩余磬折形XWV等于C。又因AE等于EB，故AN等于NB，即等于LP。加EO，则AO等于磬折形VWX，而磬折形等于C，所以AO等于C。