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Euclid: Elementa
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Proposition 44. 
PROBL. 12. PROPOS. 44. 
第四十四題 
To a given straight line to apply, in a given rectilineal angle, a parallelogram equal to a given triangle. 
AD datam rectam lineam, dato triangulo æquale parallelogrammum applicare in dato angulo rectilineo. 
一直線上。求作平行方形。與所設三角形等。而方形角、有與所設角等。 
Let AB be the given straight line, C the given triangle and D the given rectilineal angle;  thus it is required to apply to the given straight line AB, in an angle equal to the angle D, a parallelogram equal to the given triangle C. 
DATA recta linea sit A, datum triangulum B, & datus angulus rectilineus C.  Oportet igitur constituere parallelogrammum æquale triangulo B, angulum habens æqualem angulo C, & unum latus æquale rectæ A. 
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Let the parallelogram BEFG be constructed equal to the triangle C, in the angle EBG which is equal to D [I. 42];  let it be placed so that BE is in a straight line with AB;  let FG be drawn through to H,  and let AH be drawn through A parallel to either BG or EF. [I. 31]  Let HB be joined.  Then, since the straight line HF falls upon the parallels AH, EF,  the angles AHF, HFE are equal to two right angles. [I. 29]  Therefore the angles BHG, GFE are less than two right angles;  and straight lines produced indefinitely from angles less than two right angles meet; [Post. 5]  therefore HB, FE, when produced, will meet.  Let them be produced and meet at K; through the point K let KL be drawn parallel to either EA or FH, [I. 31]  and let HA, GB be produced to the points L, M.  Then HLKF is a parallelogram, HK is its diameter,  and AG, ME are parallelograms, and LB, BF the so-called complements, about HK;  therefore LB is equal to BF. [I. 43]  But BF is equal to the triangle C; therefore LB is also equal to C. [C.N. 1]  And, since the angle GBE is equal to the angle ABM, [I. 15]  while the angle GBE is equal to D, the angle ABM is also equal to the angle D. 
Constituatur triangulo B, æquale parallelogrammum D E F G, habens angulum E F G, angulo C, æqualem,    producaturque G F, ad H, ut F H, sit æqualis rectæ A,  & per H, ducatur H I, parallela ipsi F E, occurrens D E, productæ in I.  Extendatur deinde ex I, per F,              diameter I F, occurrens rectæ D G, productæ in K; & per K, ducatur K L, parallela ipsi G H, secans I H, protractam in L, producaturque E F, ad M.        Dico parallelogrammum L M F H, esse id, quod quæritur.  Habet enim latus F H, æquale datæ rectæ A, & angulum H F M, angulo dato C, æqualem,  cum angulus H F M, æqualis sit angulo E F G, qui factus est æqualis angulo C: 
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Therefore the parallelogram LB equal to the given triangle C has been applied to the given straight line AB, in the angle ABM which is equal to D.  Q. E. F. 
Denique parallelogrammum L M F H, æquale est triangulo B, cum æquale sit complemento D E F G, quod factum est æquale triangulo B. Ad datam igitur rectam lineam dato triangulo, &c.  Quod erat faciendum.

SCHOLION
QVOD si quis optet, lineam ipsam A, datam, esse unum latus parallelogrammi, non difficile erit transferre parallelogrammum F M L H, ad rectam A, ex iis, quæ in scholio propositio 31. huius liber docuimus. Si enim in N, extremitate rectæ A, fiat angulus æqualis angulo M F H, & sumatur recta N O, æqualis rectæ F M, compleaturqúe parallelogrammum, ceu in dicto scholio traditum fuit, effectum erit, quod quæritur.
ADDIT hic aliud problema Peletarius, hoc modo.

AD datam rectam lineam, dato parallelogrammo constituere æquale triangulum in dato angulo rectilineo.
SIT data recta A B; datum parallelogrammum C D E F, & datus angulus L. Producatur C D, ad G, ut D G, æqualis sit ipsi C D, & iungatur G E, recta: Eritque triangulum C E G, parallelogrammo C D E F, æquale, ut demonstrauimus scholio propos. 41. Fiat iam super data recta A B, parallelogrammum A B H I, æquale triangulo C E G, hoc est, parallelogrammo C D E F, habens angulum A, angulo L, æqualem; & producatur A I, ad K, ut sit I K, æqualis ipsi A I, iungaturque recta B K. Dico triangulum A B K, constitutum super datam rectam A B, habensque angulum A, æqualem dato angulo L; æquale esse dato parallelogrammo C D E E. Cum enim triangulum A B K, æquale sit parallelogrammo A B H I, ex scholio propos. 41. quod æquale est constructum parallelogrammo C D E F, constat propositum. 
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