Elements of Geometry: Containing the First Six Books of Euclid, with a Supplement on the Quadrature of the Circle and the Geometry of Solids ; to which are Added Elements of Plane and Spherical Trigonometry |
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Αποτελέσματα 11 - 15 από τα 79.
Σελίδα 65
2BE + 2AE ) ; and for the same reason , CD + BC2 = 2BE2 + 2ECI = 2BE3 + 2AE because EC = AE . Therefore ABS + AD : + DC + BC : = , = 4BE3 + 4AE . But 4BE ? = BD , and 4AE : = AC : ( 2. Cor . 8. 2. ) because BD and AC are both bisected ...
2BE + 2AE ) ; and for the same reason , CD + BC2 = 2BE2 + 2ECI = 2BE3 + 2AE because EC = AE . Therefore ABS + AD : + DC + BC : = , = 4BE3 + 4AE . But 4BE ? = BD , and 4AE : = AC : ( 2. Cor . 8. 2. ) because BD and AC are both bisected ...
Σελίδα 71
... CEF , the two sides BE , EF are equal to A the two CE , EF ; but the angle BEF is B greater than the angle CEF ; therefore the base BF is greater ( 24. 1. ) than the base FC ; for the same reason , CF is greater than GF .
... CEF , the two sides BE , EF are equal to A the two CE , EF ; but the angle BEF is B greater than the angle CEF ; therefore the base BF is greater ( 24. 1. ) than the base FC ; for the same reason , CF is greater than GF .
Σελίδα 76
For the same reason , CD is double of CG : But AB is equal to CD ; E therefore AF is equal to CG : And be . cause AE is equal to EC , the square of AE is equal to the square of EC : Now B D the squares of AF , FÈ are equal ( 47. 1. ) ...
For the same reason , CD is double of CG : But AB is equal to CD ; E therefore AF is equal to CG : And be . cause AE is equal to EC , the square of AE is equal to the square of EC : Now B D the squares of AF , FÈ are equal ( 47. 1. ) ...
Σελίδα 80
to the angles DAB , DBA ; therefore also the angle BDE is double of the angle DAB : For the same reason , the angle EDC is double of the angle DAC : Therefore the whole angle BDC is double of the whole angle BAC .
to the angles DAB , DBA ; therefore also the angle BDE is double of the angle DAB : For the same reason , the angle EDC is double of the angle DAC : Therefore the whole angle BDC is double of the whole angle BAC .
Σελίδα 81
Draw AF to the centre , and produce B D to C , and join CE : Therefore the segment BADC is greater than a semicircle ; F and the angles in it BAC , BEC are equal , by the first case : For the same reason , because CBED is greater than a ...
Draw AF to the centre , and produce B D to C , and join CE : Therefore the segment BADC is greater than a semicircle ; F and the angles in it BAC , BEC are equal , by the first case : For the same reason , because CBED is greater than a ...
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ABC is equal ABCD altitude angle ABC angle BAC arch base bisected Book called centre circle circle ABC circumference coincide common cylinder definition demonstrated described diameter difference divided double draw drawn equal equal angles equiangular Euclid exterior angle extremity fall fore four fourth given given straight line greater half inscribed interior join less Let ABC magnitudes manner meet multiple opposite parallel parallelogram pass perpendicular plane polygon prism produced PROP proportionals proposition proved radius ratio reason rectangle contained rectilineal figure right angles segment shewn sides similar sine solid square straight line taken tangent THEOR thing third touches triangle ABC wherefore whole
Πληροφορίες βιβλιογραφίας
| Τίτλος | Elements of Geometry: Containing the First Six Books of Euclid, with a Supplement on the Quadrature of the Circle and the Geometry of Solids ; to which are Added Elements of Plane and Spherical Trigonometry |
| Συγγραφέας | John Playfair |
| Εκδότης | E. Duyckinck, and George Long, 1824 |
| Πρωτότυπο από | Πανεπιστήμιο της Ιντιάνα |
| Ψηφιοποιήθηκε στις | 15 Ιαν. 2009 |
| Μέγεθος | 333 σελίδες |
| Εξαγωγή αναφοράς | BiBTeX EndNote RefMan |