 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 - 1765 - 464 pagesFull view - About this book
 | Euclides - 1826 - 226 pages
...manner it may be demonstrated that each of the sides FL, FM, FG, is equal to each of the sides FH, FK ; therefore the five right lines FG, FH, FK, FL, FM, are equal to one another. Wherefore with centre F and distance any one of them FG, FH, FK, FL, FM, the circle described will both pass... | |
 | Euclid - 1826 - 234 pages
...manner it may be demonstrated that each of the sides FL, FM, FG, is equal to each of the sides FH, FK ; therefore the five right lines FG, FH, FK, FL, FM, are equal to one another. Wherefore with centre F and distance any one of them FO, FH, FK, FL, FM, the circle described will both pass... | |
 | Robert Simson - Trigonometry - 1827 - 546 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... | |
 | Euclid - 1835 - 540 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... | |
 | Andrew Bell - Euclid's Elements - 1837 - 290 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... | |
 | Euclides - Geometry - 1841 - 378 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 - Euclid's Elements - 1842 - 332 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... | |
 | Euclides - 1845 - 546 pages
...perpendicular FH is equal to the perpendicular FA": 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... | |
 | Euclid - Geometry - 1845 - 218 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... | |
 | Euclides - 1846 - 292 pages
...each to each, and therefore FH is equal to FK : And in like 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, and the circle described from the centre... | |
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