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" 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 183
by Euclid - 1765 - 464 pages
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Euclid's Elements of Geometry: From the Latin Translation of Commandine. To ...

John Keill - Logarithms - 1723 - 444 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,...
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Euclid's Elements of Geometry,: From the Latin Translation of Commandine. To ...

Euclid, John Keill - Geometry - 1733 - 444 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,...
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Euclid's Elements of Geometry: From the Latin Translation of Commandine. To ...

John Keill - Geometry - 1772 - 462 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,...
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The First Six Books: Together with the Eleventh and Twelfth

Euclid - 1781 - 552 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...
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The Elements of Euclid: The Errors, by which Theon, Or Others, Have Long Ago ...

Robert Simson - Trigonometry - 1781 - 534 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...
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Elements of Geometry;: Containing the First Six Books of Euclid, with Two ...

Euclid, John Playfair - Euclid's Elements - 1795 - 462 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...
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The Elements of Euclid: Also the Book of Euclid's Data ...cor. viz. The ...

Robert Simson - Trigonometry - 1804 - 530 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 de~ fcribed from the...
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The Elements of Euclid: Viz. the First Six Books, Together with the Eleventh ...

Euclides - 1816 - 588 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...
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Elements of Geometry: Containing the First Six Books of Euclid, with a ...

John Playfair - Circle-squaring - 1819 - 350 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...
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A Popular Course of Pure and Mixed Mathematics ...: With Tables of ...

Peter Nicholson - Mathematics - 1825 - 1046 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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