Elements of Geometry, Containing the First Six Books of EuclidBaldwin, Cradock, and Joy, 1826 - 180 sider |
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Side xi
... 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 . Next in order comes ...
... 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 . Next in order comes ...
Side xii
... 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 at first sight appeared easy ; but the mistake was soon perceived , and all ...
... 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 at first sight appeared easy ; but the mistake was soon perceived , and all ...
Side xiii
... 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 simple a way as we trace a circle with the compasses ...
... 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 simple a way as we trace a circle with the compasses ...
Side xiv
... problem of the trisection of an angle , which is of the same kind with that of doubling the cube , both of which ... problems with the rule and compasses only , that they could not be persuaded to give it up ; they made many fruitless ...
... problem of the trisection of an angle , which is of the same kind with that of doubling the cube , both of which ... problems with the rule and compasses only , that they could not be persuaded to give it up ; they made many fruitless ...
Side xvi
... problems in question , so as to approximate the truth near enough for practical purposes ; most of these methods are now lost , but those of four eminent geometricians , viz . Dinostratus , Nicomedes , Pappus , and Diocles , deserve ...
... problems in question , so as to approximate the truth near enough for practical purposes ; most of these methods are now lost , but those of four eminent geometricians , viz . Dinostratus , Nicomedes , Pappus , and Diocles , deserve ...
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Vanlige uttrykk og setninger
ABC is equal adjacent angles Algebra angle ABC angle ACB angle BAC angles equal base BC bisected centre circle ABC circum circumference BC diameter double draw equal angles equal circles equal right lines equal to F equi equiangular equimultiples Euclid EUCLID'S ELEMENTS exceed exterior angle fore four magnitudes fourth Geometry given circle given point given right line gnomon greater ratio hence inscribed join less Let ABC multiple parallel parallelogram perpendicular polygon proportional Q. E. D. Deduction Q. E. D. PROPOSITION rectangle contained remaining angle right angles right line AB right line AC sector HEF segment side BC similar and similarly square of AC subtending THEOREM tiple touches the circle triangle ABC triangle DEF whence whole
Populære avsnitt
Side xxvi - 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.
Side 74 - The straight line drawn at right angles to the diameter of a circle, from the extremity of it, falls without the circle...
Side 33 - The complements of the parallelograms, which are about the diameter of any parallelogram, are equal to one another.
Side 148 - 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 27 - 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.
Side 8 - 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.
Side 73 - 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.
Side 99 - 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...
Side 7 - ... 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.
Side 88 - From a given circle to cut off a segment, which shall contain an angle equal to a given rectilineal angle.