| Robert Simson - Trigonometry - 1762 - 488 pages
...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 - Trigonometry - 1775 - 534 pages
...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 pages
...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 - Trigonometry - 1804 - 530 pages
...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 pages
...• 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 - Euclid's Elements - 1828 - 542 pages
...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 pages
...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 pages
...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 - Geometry - 1837 - 410 pages
...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 - Geometry - 1838 - 434 pages
...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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