The Elements of Euclid: The Errors, by which Theon, Or Others, Have Long Ago Vitiated These Books, are Corrected and Some of Euclid's Demonstrations are Restored. Also, to this Second Edition is Added the Book of Euclid's Data. In Like Manner Corrected. viz. the first six books, together with the eleventh and twelfth |
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Resultat 1-5 av 10
Side 209
... and find its center X , which will either be within c . s .. the triangle , or in one of
its fides , or without it . First , Let the center X be within the triangle , and join LX ,
MX , NX . AB is greater than LX . if not , AB must either be equal to ; or less than ...
... and find its center X , which will either be within c . s .. the triangle , or in one of
its fides , or without it . First , Let the center X be within the triangle , and join LX ,
MX , NX . AB is greater than LX . if not , AB must either be equal to ; or less than ...
Side 265
... lines to the extremities of the circumferences , and upon each of the triangles
thus made erecting pyramids having the fame vertex with the cone , and so on ,
there must at length remain some fegments of the cone which are together lessb
b .
... lines to the extremities of the circumferences , and upon each of the triangles
thus made erecting pyramids having the fame vertex with the cone , and so on ,
there must at length remain some fegments of the cone which are together lessb
b .
Side 293
... yet it is science which must give the reason why two sides of a triangle are
greater than the third . but the right anfwer to ... is , that the number of Axioms
ought not to be encreased without necessity , as it must be if these Propositions
be not ...
... yet it is science which must give the reason why two sides of a triangle are
greater than the third . but the right anfwer to ... is , that the number of Axioms
ought not to be encreased without necessity , as it must be if these Propositions
be not ...
Side 294
Book I. that these circles must meet one another , because FD and GH are
together greater than FG ? and this determination is easier to be understood than
that which Mr. Thomas Simpson derives from it , and puts instead of Euclid's , in
the ...
Book I. that these circles must meet one another , because FD and GH are
together greater than FG ? and this determination is easier to be understood than
that which Mr. Thomas Simpson derives from it , and puts instead of Euclid's , in
the ...
Side 300
... by the help of this Definition , and some of the preceeding Propositions . and
because in the Demonstration , this Proposition must be brought in , viz . straight
lines from the center of a circle to the circumference are equal , and that the point
...
... by the help of this Definition , and some of the preceeding Propositions . and
because in the Demonstration , this Proposition must be brought in , viz . straight
lines from the center of a circle to the circumference are equal , and that the point
...
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The Elements of Euclid, Viz: The Errors, by which Theon, Or Others, Have ... Robert Simson Uten tilgangsbegrensning - 1775 |
The Elements of Euclid: The Errors by which Theon, Or Others, Have Long ... Robert Simson Uten tilgangsbegrensning - 1827 |
The Elements of Euclid: The Errors, by which Theon, Or Others, Have Long Ago ... Robert Simson Uten tilgangsbegrensning - 1781 |
Vanlige uttrykk og setninger
added alſo altitude angle ABC angle BAC baſe BC is given becauſe Book Book XI caſe circle circle ABCD circumference common cone cylinder Definition demonſtrated deſcribed diameter divided double draw drawn equal equal angles equiangular equimultiples exceſs fame fides firſt folid fore four fourth given angle given in poſition given in ſpecies given magnitude given ratio given ſtraight line greater Greek half join leſs likewiſe magnitude manner meet muſt oppoſite parallel parallelogram perpendicular plane priſm produced PROP proportionals Propoſition pyramid reaſon rectangle remaining right angles ſame ſecond ſegment ſhall ſhewn ſides ſimilar ſolid ſquare ſquare of AC taken THEOR theſe third thro triangle ABC wherefore whole
Populære avsnitt
Side 5 - Let it be granted that a straight line may be drawn from any one point to any other point.
Side 163 - If two triangles have one angle of the one equal to one angle of the other and the sides about these equal angles proportional, the triangles are similar.
Side 48 - If a straight line be divided into any two parts, the squares of the whole line, and of one of the parts, are equal to twice the rectangle contained by the whole and that part, together with the square of the other part. Let the straight line AB be divided into any two parts in the point C; the squares of AB, BC are equal to twice the rectangle AB, BC, together with the square of AC.
Side 73 - The straight line drawn at right angles to the diameter of a circle, from the extremity of it, falls without the circle; and no straight line can be drawn from the extremity between that straight line and the circumference, so as not to cut the circle...
Side 105 - If two triangles have two angles of the one equal to two angles of the other, each to each, and one side equal to one side, viz. either the sides adjacent to the equal...
Side 3 - A circle is a plane figure contained by one line, which is called the circumference, and is such that all straight lines drawn from a certain point within the figure to the circumference, are equal to one another.
Side 167 - Similar triangles are to one another in the duplicate ratio of their homologous sides.
Side 54 - AB be the given straight line ; it is required to divide it into two parts, so that the rectangle contained by the whole, and one of the parts, shall be equal to the square of the other part.
Side 47 - If a straight line be divided into two equal parts, and also into two unequal parts; the rectangle contained by the unequal parts, together with the square of the line between the points of section, is equal to the square of half the line.
Side 37 - To describe a parallelogram that shall be equal to a given triangle, and have one of its angles equal to a given rectilineal angle.