To a given straight line to apply a parallelogram equal to a given rectilineal figure and deficient by a parallelogrammic figure similar to a given one : thus the given rectilineal figure must not be greater than the parallelogram described on the half of the straight line and similar to the defect.
在给定线段上作一个平行四边形，使其等于给定直线形，且亏缺一个与给定平行四边形相似的平行四边形；但给定直线形不能大于以线段一半为底且与亏形相似的平行四边形。
在直线 AB 上贴合与图形 C 等积、亏形与已知平行四边形 D 相似的平行四边形：经过命题 25、26 的辅助点 E～W 完成贴合构造。
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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, not being greater than the parallelogram described on the half of AB and similar to the defect, and D the parallelogram to which the defect is required to be similar; thus it is required to apply to the given straight line AB a parallelogram equal to the given rectilineal figure C and deficient by a parallelogrammic figure which is similar to D. Let AB be bisected at the point E, and on EB let EBFG be described similar and similarly situated to D; [VI. 18] let the parallelogram AG be completed. If then AG is equal to C, that which was enjoined will have been done; for there has been applied to the given straight line AB the parallelogram AG equal to the given rectilineal figure C and deficient by a parallelogrammic figure GB which is similar to D. But, if not, let HE be greater than C. Now HE is equal to GB; therefore GB is also greater than C.
设AB为给定线段，C为给定直线形，D为给定平行四边形。取AB中点E，在EB上作平行四边形EBFG与D相似且同位，补全平行四边形AG。
Let KLMN be constructed at once equal to the excess by which GB is greater than C and similar and similarly situated to D. [VI. 25] But D is similar to GB; therefore KM is also similar to GB. [VI. 21] Let, then, KL correspond to GE, and LM to GF. Now, since GB is equal to C, KM, therefore GB is greater than KM; therefore also GE is greater than KL, and GF than LM.
若AG等于C，则完成；否则，设HE大于C。因HE等于GB，故GB大于C。作平行四边形KLMN等于GB与C的差，且与D相似同位。
Let GO be made equal to KL, and GP equal to LM; and let the parallelogram OGPQ be completed; therefore it is equal and similar to KM. Therefore GQ is also similar to GB; [VI. 21] therefore GQ is about the same diameter with GB. [VI. 26] Let GQB be their diameter, and let the figure be described. Then, since BG is equal to C, KM, and in them GQ is equal to KM, therefore the remainder, the gnomon UWV, is equal to the remainder C.
因D与GB相似，故KM与GB相似。设KL对应GE，LM对应GF。因GB大于KM，故GE大于KL，GF大于LM。取GO等于KL，GP等于LM，完成平行四边形OGPQ，则其等于且相似于KM。
And, since PR is equal to OS, let QB be added to each; therefore the whole PB is equal to the whole OB. But OB is equal to TE, since the side AE is also equal to the side EB; [I. 36] therefore TE is also equal to PB. Let OS be added to each; therefore the whole TS is equal to the whole, the gnomon VWU. But the gnomon VWU was proved equal to C; therefore TS is also equal to C.
因GQ与GB相似且同对角线，设对角线为GQB。因BG等于C与KM之和，且GQ等于KM，故剩余磬折形UWV等于C。又因PR等于OS，加QB得PB等于OB，而OB等于TE，故TE等于PB。加OS得TS等于磬折形VWU，即等于C。