Elements of Geometry, Containing the First Six Books of EuclidBaldwin, Cradock, and Joy, 1826 - 180 σελίδες |
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Σελίδα 178
... sector GBC to the sector HEF . Take any number of circumferences CK , KL , each equal to BC , and any number whatever FM , MN , each equal to EF , and join GK , GL , HM , HN . And because the A circumferences BC , CK , KL , are equal to ...
... sector GBC to the sector HEF . Take any number of circumferences CK , KL , each equal to BC , and any number whatever FM , MN , each equal to EF , and join GK , GL , HM , HN . And because the A circumferences BC , CK , KL , are equal to ...
Σελίδα 179
... sector GBC is equal to the whole sector GCK . For the same reason the sector GKL is equal to each of them GKC , GCB ; hence the three sec- tors GBC , GCK , GKL , are equal to one another . For the same reason also the sectors HEF , HFM ...
... sector GBC is equal to the whole sector GCK . For the same reason the sector GKL is equal to each of them GKC , GCB ; hence the three sec- tors GBC , GCK , GKL , are equal to one another . For the same reason also the sectors HEF , HFM ...
Σελίδα 180
Euclid. sector GBL will also exceed the sector HEN ; if equal , equal ; and if less , less ; therefore as the circumference BC is to EF so is the sector GBC to the sector HEF . Therefore in equal circles , & c . Q. E. D. * Deduction ...
Euclid. sector GBL will also exceed the sector HEN ; if equal , equal ; and if less , less ; therefore as the circumference BC is to EF so is the sector GBC to the sector HEF . Therefore in equal circles , & c . Q. E. D. * Deduction ...
Σελίδα 179
... sector GBC is equal to the whole sector GCK . For the same reason the sector GKL is equal to each of them GKC , GCB ; hence the three sec- tors GBC , GCK , GKL , are equal to one another . For the same reason also the sectors HEF , HFM ...
... sector GBC is equal to the whole sector GCK . For the same reason the sector GKL is equal to each of them GKC , GCB ; hence the three sec- tors GBC , GCK , GKL , are equal to one another . For the same reason also the sectors HEF , HFM ...
Σελίδα
... sector GBC to the sector HEF . C D K E F b M Take any number of circumferences CK , KL , each equal to BC , and any number whatever FM , MN , each equal to EF , and join GK , GL , HM , HN . And because the circumferences BC , ск , KL ...
... sector GBC to the sector HEF . C D K E F b M Take any number of circumferences CK , KL , each equal to BC , and any number whatever FM , MN , each equal to EF , and join GK , GL , HM , HN . And because the circumferences BC , ск , KL ...
Άλλες εκδόσεις - Προβολή όλων
Elements of Geometry; Containing the First Six Books of Euclid ... Euclid,John Playfair Πλήρης προβολή - 1814 |
Συχνά εμφανιζόμενοι όροι και φράσεις
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
Δημοφιλή αποσπάσματα
Σελίδα 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.
Σελίδα 74 - The straight line drawn at right angles to the diameter of a circle, from the extremity of it, falls without the circle...
Σελίδα 33 - The complements of the parallelograms, which are about the diameter of any parallelogram, are equal to one another.
Σελίδα 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.
Σελίδα 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.
Σελίδα 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.
Σελίδα 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.
Σελίδα 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...
Σελίδα 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.
Σελίδα 88 - From a given circle to cut off a segment, which shall contain an angle equal to a given rectilineal angle.