Elements of Geometry, Containing the First Six Books of Euclid |
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Seite xi
... have equal areas : a problem not easy of solution , and shows that Geometry must have then made a great progress . The ingenious theory of the five regular bodies originated also about the same time in the Pytha- gorean school .
... have equal areas : a problem not easy of solution , and shows that Geometry must have then made a great progress . The ingenious theory of the five regular bodies originated also about the same time in the Pytha- gorean school .
Seite xii
honour among the list of Geometers , who at- tempted to solve the celebrated problem of the duplication of the cube , which at that period began to be pursued with ardour . The circumstance of this problem is well known ; its solution ...
honour among the list of Geometers , who at- tempted to solve the celebrated problem of the duplication of the cube , which at that period began to be pursued with ardour . The circumstance of this problem is well known ; its solution ...
Seite xiii
... learned applications of the same theory to the problem of the duplication of the cube ; and from the result of his labours , it appears that if we possessed the means of de- scribing conic sections by one continued motion , in as ...
... learned applications of the same theory to the problem of the duplication of the cube ; and from the result of his labours , it appears that if we possessed the means of de- scribing conic sections by one continued motion , in as ...
Seite xiv
I cannot , however , pass over the problem of the trisection of an angle , which is of the same kind with that of doubling the cube , both of which were equally agitated in the school of Plato ; and although a solution was not to be ...
I cannot , however , pass over the problem of the trisection of an angle , which is of the same kind with that of doubling the cube , both of which were equally agitated in the school of Plato ; and although a solution was not to be ...
Seite xvi
researches they were continually making ; new theories were introduced , and some ingenious in- struments for solving the two problems in question , so as to approximate the truth near enough for practical purposes ; most of these ...
researches they were continually making ; new theories were introduced , and some ingenious in- struments for solving the two problems in question , so as to approximate the truth near enough for practical purposes ; most of these ...
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ABC is equal ABCD alternate angle ABC angle ACB angle BAC angles equal applied base base BC bisected centre circle ABC circumference common consequently demonstrated described diameter divided double draw equal equal angles equiangular equilateral equimultiples exceed extreme fall figure fore four fourth Geometry given circle given rectilineal given right line greater half hence inscribed join less Let ABC magnitudes manner mean meet multiple opposite parallel parallelogram pass perpendicular PROBLEM produced proportional PROPOSITION ratio reason rectangle rectangle contained rectilineal figure remaining angle right angles right line AC sector segment shown sides similar square taken THEOREM third triangle ABC whence Wherefore whole
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Seite xxiv - A diameter of a circle is a straight line drawn through the centre, and terminated both ways by the circumference. XVIII. A semicircle is the figure contained by a diameter and the part of the circumference cut off by the diameter. XIX. "A segment of a circle is the figure contained by a straight line, and the circumference it cuts off.
Seite 72 - The straight line drawn at right angles to the diameter of a circle, from the extremity of it, falls without the circle...
Seite 31 - The complements of the parallelograms, which are about the diameter of any parallelogram, are equal to one another.
Seite 146 - 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.
Seite 25 - And because the angle ABC is equal to the angle BCD, and the angle CBD to the angle ACB, therefore the whole angle ABD is equal to the whole angle ACD • (ax.
Seite 6 - To bisect a given finite straight line, that is, to divide it into two equal parts. Let AB be the given straight line : it is required to divide it intotwo equal parts.
Seite 71 - DH; (I. def. 15.) therefore DH is greater than DG, the less than the greater, which is impossible : therefore no straight line can be drawn from the point A, between AE and the circumference, which does not cut the circle : or, which amounts to the same thing, however great an acute angle a straight line makes with the diameter at the point A, or however small an angle it makes with AE, the circumference must pass between that straight line and the perpendicular AE.
Seite 97 - To describe a square about a given circle. Let ABCD be the given circle ; it is required to describe a square about it. . Draw two diameters AC, BD of the circle ABCD, at right angles to one another, and through the points A, B, • 17.3. C, D, draw...
Seite 5 - ... equal to them, of the other. Let ABC, DEF be two triangles which have the two sides AB, AC equal to the two sides DE, DF, each to each, viz. AB equal to DE, and AC to DF ; but the base CB greater than the base EF ; the angle BAC is likewise greater than the angle EDF.
Seite 86 - From a given circle to cut off a segment, which shall contain an angle equal to a given rectilineal angle.