Elements of Geometry: Containing the First Six Books of Euclid, with a Supplement on the Quadrature of the Circle and the Geometry of Solids ; to which are Added Elements of Plane and Spherical Trigonometry |
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Página 21
And that a circle may be described from any centre , at any distance from that centre . AXIOMS . I. THINGS which are equal to the same thing are equal to one another . II . If equals be added to equals , the wholes are equal . III .
And that a circle may be described from any centre , at any distance from that centre . AXIOMS . I. THINGS which are equal to the same thing are equal to one another . II . If equals be added to equals , the wholes are equal . III .
Página 22
... AK CB are equal to one another ; and the triangle ABC is therefore equilateral , and it is described upon the given straight line AB . Which was required to be done . PROP . II . PROB . From a given point to draw a straight line ...
... AK CB are equal to one another ; and the triangle ABC is therefore equilateral , and it is described upon the given straight line AB . Which was required to be done . PROP . II . PROB . From a given point to draw a straight line ...
Página 48
... equal to the given angle D : Wherefore there has been described a parallelogram FECG equal to a given triangle ABC , haring one of its angles CEF equal to the given angle D. Which was to be done . 1 PROP . XLIII . THEOR .
... equal to the given angle D : Wherefore there has been described a parallelogram FECG equal to a given triangle ABC , haring one of its angles CEF equal to the given angle D. Which was to be done . 1 PROP . XLIII . THEOR .
Página 50
... figure ABCD is equal to the whole parallelogram KFLM ; therefore the parallelogram KFLM has been described equal to the given rectilineal figure ABCD , having the angle FKM equal to the given angle E. Which was to be done . Cor .
... figure ABCD is equal to the whole parallelogram KFLM ; therefore the parallelogram KFLM has been described equal to the given rectilineal figure ABCD , having the angle FKM equal to the given angle E. Which was to be done . Cor .
Página 51
therefore each of the opposite angles ABE , BED is a right an- А B gle ; wherefore the figure ADEB is rectangular , and it has been demonstrated that it is equilateral ; it is therefore a square , and it is described upon the given ...
therefore each of the opposite angles ABE , BED is a right an- А B gle ; wherefore the figure ADEB is rectangular , and it has been demonstrated that it is equilateral ; it is therefore a square , and it is described upon the given ...
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ABC is equal ABCD altitude angle ABC angle BAC arch base bisected Book called centre circle circle ABC circumference coincide common cylinder definition demonstrated described diameter difference divided double draw drawn equal equal angles equiangular Euclid exterior angle extremity fall fore four fourth given given straight line greater half inscribed interior join less Let ABC magnitudes manner meet multiple opposite parallel parallelogram pass perpendicular plane polygon prism produced PROP proportionals proposition proved radius ratio reason rectangle contained rectilineal figure right angles segment shewn sides similar sine solid square straight line taken tangent THEOR thing third touches triangle ABC wherefore whole