Elements of Geometry, Containing the First Six Books of EuclidBaldwin, Cradock, and Joy, 1826 - 180 Seiten |
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Seite 38
... reason the triangle KGB is equal to the tri- angle KEB . Therefore the triangle BEK is equal to the triangle BGK , and the triangle HDK to the triangle DKF ; and the triangle BEK , together with the triangle HDK , will be equal to the ...
... reason the triangle KGB is equal to the tri- angle KEB . Therefore the triangle BEK is equal to the triangle BGK , and the triangle HDK to the triangle DKF ; and the triangle BEK , together with the triangle HDK , will be equal to the ...
Seite 41
... reason AB , AH , are in one and the same right line . And because the angle DBC is equal to the angle FBA , each of them is a right angle : add ABC , which is common : therefore the whole angle DBA is equal to the whole angle FBC.c But ...
... reason AB , AH , are in one and the same right line . And because the angle DBC is equal to the angle FBA , each of them is a right angle : add ABC , which is common : therefore the whole angle DBA is equal to the whole angle FBC.c But ...
Seite 48
... reason HF is also a square made upon HG , that is , equal to the square of AC . Wherefore HF and сK are the squares of AC and CB , and because the rectangle AG is equal to the rectangle GE , and AG is that contained under AC and CB ...
... reason HF is also a square made upon HG , that is , equal to the square of AC . Wherefore HF and сK are the squares of AC and CB , and because the rectangle AG is equal to the rectangle GE , and AG is that contained under AC and CB ...
Seite 55
... reason are also CEB , CBE , each of them half right angles . Therefore the whole angle AEB is a right angle . And because the angle GEF is half a right angle , and EGF a right angle , for it is equal to the 29. 1 . interior and opposite ...
... reason are also CEB , CBE , each of them half right angles . Therefore the whole angle AEB is a right angle . And because the angle GEF is half a right angle , and EGF a right angle , for it is equal to the 29. 1 . interior and opposite ...
Seite 69
... reason CF is also greater than FG . Again , because GF , FE , are greater b b 20. 1 . than GE , but GE is equal to ED ; GF , FE , will be greater than ED , take away the common part FE ; therefore the remainder GF is greater than the ...
... reason CF is also greater than FG . Again , because GF , FE , are greater b b 20. 1 . than GE , but GE is equal to ED ; GF , FE , will be greater than ED , take away the common part FE ; therefore the remainder GF is greater than the ...
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Häufige Begriffe und Wortgruppen
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
Beliebte Passagen
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.