 | Robert Simson - 1762 - 488 σελίδες
...multiple of F, that CD is of F ; wherefore : KH i& equal to CD". take away the common magnitude CH, then the remainder KC is equal to the remainder HD. but KC is equal to F, HD therefore is equal to F. But let GB be a multiple of E ; then HD is the fame multiple of F. Make... | |
 | Robert Simson - 1775 - 534 σελίδες
...of F, that CD is of F ; wherefore KH is equal to CD • : Take a' way the common magnitude CH, then the remainder KC is equal to the remainder HD : But KC is equal to F ; HD therefore is equal to F. But let GB be a multiple of E ; then „ HD is the farne multiple of... | |
 | Euclid - 1781 - 552 σελίδες
...CD is of F; • i. At. 5. wherefore KH is equal to CD • : Take away the common magnitude CH, then the remainder KC is equal to the remainder HD : But KC is equal to F ; HD therefore is equal to F. But let GB be a multiple of E ; then HD is the fame multiple ot F : Make... | |
 | Robert Simson - 1804 - 530 σελίδες
...that CD is of F ; wherefore 1. *. Ax. 5. KH is equal to CD a. take away the common magnitude CH, then the remainder KC is equal to the remainder HD. but KC is equal to F, HD therefore is equal to F. ' But let GB be a multiple E; then HD is the fame multiple of F, Make CK... | |
 | Euclides - 1816 - 588 σελίδες
...• 1 Ax. s. is of F; wherefore KH is equal to CD a: 15 3) KF Take away the common magnitude CH, then the remainder KC is equal to the remainder HD: But KC is equal to F ; HD therefore is equal to F. But let GB be a multiple of E; then v HD is the same multiple of F: Make... | |
 | Euclid, Dionysius Lardner - 1828 - 542 σελίδες
...CD is of F : wherefore KH is equal (Ax. I.) to CD : take away BD k K the common magnitude CH, then the remainder KC is equal to the remainder HD : but KC is equal (const.) to F : therefore HD is equal to F. Next let GB be a multiple of E ; HD shall be the same multiple... | |
 | Euclides - 1834 - 518 σελίδες
...multiple of F, that CD is of F i wherefore KH is equal * to CD : take away the common magnitude CH, then the remainder KC is equal to the remainder HD : but KC is equal t | Constr. to F ; therefore HD is equal to F. Next, let GB be a multiple of E : HD shall be the same... | |
 | Euclid - 1835 - 540 σελίδες
...Talrn •" ^ K a 1. Ax. 5. wherefore, KH is equal to CD a : Take away the common magnitude CH, then the remainder KC is equal to the remainder HD: but KC is equal to F, HD therefore is equal to F. Next, Let GB be a multiple of E ; then will HD be the same multiple of... | |
 | Euclid, James Thomson - 1837 - 410 σελίδες
...same multiple of F, that CD is of F; wherefore (V. ax. 1.) KH is equal to CD. Take away ('1 1, then the remainder KC is equal to the remainder HD : but KC is equal to F; HD thereforeis equal to F. But let GB be a multiple of E ; then HD is the same multiple of F. Make... | |
 | Robert Simson - 1838 - 434 σελίδες
...that CD is of F ; wherefore KH is equal to CD (1. Ax. 5.) : take away the common magnitude CH, then the remainder KC is equal to the remainder HD ; but KC is equal to F; HD therefore is equal to F. But let GB be a multiple of E : then HD is the same multiple of F : make... | |
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GB is equal to E, and CK to F ; therefore AB is the same multiple of E, that KH is of F: But AB... 
