FK : in the same manner it may be demonstrated, that FL, FM, FG are each of them equal to FH, or FK : therefore the five straight lines FG, FH, FK, FL, FM are equal to one another : wherefore the circle described from the centre F, at the distance of... Euclid's Elements of Geometry: The First Six, the Eleventh and Twelfth Books - Page 183by Euclid, Edmund Stone - 1765 - 464 pagesFull view - About this book
| John Keill - Logarithms - 1723 - 364 pages
...Perpendicular F K. In the fame Manner we demonstrate, that FL, FM,- or FG, is equal to FH, or F K. **Therefore the five Right Lines FG, FH, FK, FL, FM, are equal to** each other. And fo a Circle defcribed on the Center F, with either of the Diftances FG, FH, FK, FL,... | |
| John Keill - Trigonometry - 1733 - 397 pages
...Perpendicular FH equal to the Perpendicular F K. In the fame Manner we demonftrate, that FL, FM, or FG, is **equal to FH, or FK. Therefore the five Right Lines FG, FH, FK, FL, FM, are equal to** each other. And fo a Circle defcribed oh the Center F, with either of the Diftances FG, FH, FK, FL,... | |
| John Keill - Logarithms - 1772 - 399 pages
...the Perpendicular F K. In the fame manner we demonftrate, that FL, FM, or FG, is equal to FH, or F K. **Therefore the five Right • Lines FG, FH, FK, FL, FM, are equal to** each other, and fo a Circle defcribed on the Centre F, with either of the Diftances FG, FH, FK, FL,... | |
| Euclid - 1781 - 520 pages
...perpendicular FH is equal to the perpendicular FK : In the fame manner it may be demonftrated that FL, **FM, FG are each of them equal to FH or FK ; therefore the five** ftraight lines FG, FH, FK, FL, FM are equal to one another : Wherefore the cifcle defcribed from the... | |
| Robert Simson - Trigonometry - 1781 - 466 pages
...perpendicular FH is equal to the perpendicular FK. in the fwne manner it may be demonftrated that FL, **FM, FG are each of them equal to FH or FK; therefore the five** ftraight lines FG, FH, FK, FL, FM are equal to one another. wherefore the circle defcribe .'. from... | |
| Euclid, John Playfair - Electronic book - 1795 - 400 pages
...perpendicular FH is equal to the perpendicular FK : in the fame manner it may be demonftrated, that FL, **FM, FG are each of them equal to FH or FK : therefore the five** ftraight lines FG, FH, FK, FL, FM are equal to one another : wherefore the circle defcribed from the... | |
| Robert Simson - Trigonometry - 1804
...perpendicular FH is equal to the perpendicular FK. in the fame manner it may be demonftrated that FL, **FM, FG are each of them equal to FH or FK; therefore the five** ftraight lines FG, FH, FK, FL, FM are equal to one another, wherefore the circle de~ fcribed from the... | |
| Euclides - 1816 - 528 pages
...perpendicular FH is equal to the perpendicular FK : In the same manner it may be demonstrated ; that FL, **FM, FG are each of them equal to FH, or FK : Therefore the five** straight lines FG, FH, FK, FL, FM are equal to one another : Wherefore the circle described from the... | |
| John Playfair - Circle-squaring - 1819 - 333 pages
...perpendicular FH is equal to the perpendicular FK : in the same manner it may be demonstrated, that FL, **FM, FG are each of them equal to FH or FK , therefore the five** straight lines FG, FH, FK, FL, FM are equal to one another : wherefore the circle described from the... | |
| Peter Nicholson - Mathematics - 1825 - 372 pages
...perpendicular FH is equal to the perpendicular FK : In the same manner it may be demonstrated, that FL, **FM, FG are each of them equal to FH or FK : Therefore the** uve straight linei FG, FH, FK , FL, FM are equal to on* another: Wherefore the circb described from... | |
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