Elements of Geometry, Containing the First Six Books of EuclidBaldwin, Cradock, and Joy, 1826 - 180 Seiten |
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Seite 5
... equal to CB ; wherefore the three , CA , AB , BC , are equal to one another ; and , consequently , the triangle ABC is equilateral , and it is described upon the given finite right line AB . Q. E. F. PROPOSITION II . PROBLEM . From a ...
... equal to CB ; wherefore the three , CA , AB , BC , are equal to one another ; and , consequently , the triangle ABC is equilateral , and it is described upon the given finite right line AB . Q. E. F. PROPOSITION II . PROBLEM . From a ...
Seite 7
... equal sides . Let there be two triangles , ABC , DEF , which have the two sides ab , ac , equal to the two sides de , df , each to each ; namely , the side AB equal to the side DE , and the side Ac equal to DF ; also the angle BAC equal ...
... equal sides . Let there be two triangles , ABC , DEF , which have the two sides ab , ac , equal to the two sides de , df , each to each ; namely , the side AB equal to the side DE , and the side Ac equal to DF ; also the angle BAC equal ...
Seite 8
... equal right lines being produced , the angles under the base shall be equal to one another . F B C G There- Let ABC be an isosceles triangle , having the side AB equal to the side Ac , and produce the right lines Ab , AC , directly ...
... equal right lines being produced , the angles under the base shall be equal to one another . F B C G There- Let ABC be an isosceles triangle , having the side AB equal to the side Ac , and produce the right lines Ab , AC , directly ...
Seite 9
... ABC will be equal to Ax . 3 . the remaining angle ACB : and they are at the base of the triangle ABC . But it has also been proved , that the angle FBC is equal to the angle GCB , which are under the base . Therefore the angles which ...
... ABC will be equal to Ax . 3 . the remaining angle ACB : and they are at the base of the triangle ABC . But it has also been proved , that the angle FBC is equal to the angle GCB , which are under the base . Therefore the angles which ...
Seite 11
... AB , AC , equal to two sides DE , DF , each to each , viz . AB equal to DE , and AC to DF ; and they have the base BC equal to the base EF . Then is the angle BAC equal to the angle EDF . For the tri- angle ABC being applied to the ...
... AB , AC , equal to two sides DE , DF , each to each , viz . AB equal to DE , and AC to DF ; and they have the base BC equal to the base EF . Then is the angle BAC equal to the angle EDF . For the tri- angle ABC being applied to the ...
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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
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Seite 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.
Seite 74 - The straight line drawn at right angles to the diameter of a circle, from the extremity of it, falls without the circle...
Seite 33 - The complements of the parallelograms, which are about the diameter of any parallelogram, are equal to one another.
Seite 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.
Seite 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.
Seite 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.
Seite 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.
Seite 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...
Seite 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.
Seite 88 - From a given circle to cut off a segment, which shall contain an angle equal to a given rectilineal angle.