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Euclid: Elementa

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Proposition 38. 
THEOR. 28. PROPOS. 38. 
第三十八題 
Triangles which are on equal bases and in the same parallels are equal to one another. 
TRIANGVLA super æqualibus basibus constituta, & in eisdem parallelis, inter se sunt æqualia. 
兩平行線內。有兩三角形。若底等。則兩形必等。 
Let ABC, DEF be triangles on equal bases BC, EF and in the same parallels BF, AD;  I say that the triangle ABC is equal to the triangle DEF. 
INTER parallelas A B, C D, & super æquales bases C E, F D, sint constituta triangula A C E, B F D.  Dico ipsa esse æqualia. 
For let AD be produced in both directions to G, H;  through B let BG be drawn parallel to CA, [I. 31] and through F let FH be drawn parallel to DE.  Then each of the figures GBCA, DEFH is a parallelogram; and GBCA is equal to DEFH;  for they are on equal bases BC, EF and in the same parallels BF, GH. [I. 36]  Moreover the triangle ABC is half of the parallelogram GBCA; for the diameter AB bisects it. [I. 34]  And the triangle FED is half of the parallelogram DEFH; for the diameter DF bisects it. [I. 34]  [But the halves of equal things are equal to one another.]  Therefore the triangle ABC is equal to the triangle DEF. 
  Ducatur enim E G, parallela ipsi A C, & D H, ipsi B F.  Eruntque parallelogramma A C E G, B F D H, æqualia.    Cum igitur horum dimidia sint triangula A C E, B E D;      erunt hæc inter se æqualia. 
Therefore etc.  Q. E. D. 
Triangula ergo super æqualibus basibus, &c.  Quod erat ostendendum.

SCHOLION
CONVERSA huius ostendetur ab Euclide propos. 40.
COLLIGITVR autem ex hac propositione, si a quouis angulo trianguli dati linea recta ducatur diuidens latus oppositum bifariam, triangulum quoque bifariam secari. Ducatur enim in triangulo A B C, ex angulo A, recta A D, diuidens bifariam latus B C, in D. Dico triangulum A B C, bifariam quoque secari. Si enim per A, ducatur parallela ipsi B C, erunt duo triangula A B D, A D C, inter easdem parallelas; & super æquales bases. Quare æqualia erunt.

EX PELETARIO
A puncto quouis dato in uno latere trianguli propositi lineam rectam ducere, quæ bifariam secet triangulum datum.
SIT triangulum A B C, & punctum datum D, in latere B C. Oportet igitur ex D, rectam lineam ducere, quæ bifariam diuidat triangulum. Quod si punctum D, diuidat latus B C, bifartam, recta D A, ducta ad A, diuidet triangulum bifariam, ut in hoc scholio est ostensum: Si vero D, non diuidit B C, bifariam, secetur B C, bifariam in E. Deinde ex D, ad angulum oppositum A ducatur recta D A, & per E, parallela E F, ipsi D A, secans A C, in F. Si igitur ducatur recta D F, erit triangulum diuisum bifariam a linea D F. Nam ducta recta E A, erunt triangula E F A, E F D, æqualia, cum sint super eandem basin E F, & inter easdem parallelas E F, A D. Addito igitur communi C F E, erunt tota triangula A E C, C D F, æqualia: Est autem A E C, dimidium totius A B C, ut iam fuit ostensum. Igitur & C D F, dimidium est eiusdem trianguli A B C. quod erat probandum.
QVOD si punctum D, fuerit in altera medietate E C, eodem modo problema conficiemus: sed tunc triangulum abscindetur ad partes B, trapezium vero ad partes C, ut figura præsens satis indicat. Demonstratio autem eadem est, si in ea mutetur litera B, in C, & C, in B. Hoc tamen problema multo nos uniuersalius proponemus ad finem sexti libri.
 
 
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