The Synoptical Euclid; Being the First Four Books of Euclid's Elements of Geometry from the Edition of Dr. Robert Simson ... With Exercises. By S. A. Good ... Second Edition |
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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 . ” XX .
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 . ” XX .
Página 62
... or GF , describe the semicircle BHF , and produce DE to H : the square described upon EH shall be equal to the given rectilineal figure A. HI B G E Join GH ; and because the straight line BF is 62 EUCLID'S ELEMENTS . B ...
... or GF , describe the semicircle BHF , and produce DE to H : the square described upon EH shall be equal to the given rectilineal figure A. HI B G E Join GH ; and because the straight line BF is 62 EUCLID'S ELEMENTS . B ...
Página 80
And this is all that is to be understood , when , in the Greek text and translations from it , the angle of the semicircle is said to be greater than any acute rectilineal angle , and the remaining angle less than any rectilineal angle ...
And this is all that is to be understood , when , in the Greek text and translations from it , the angle of the semicircle is said to be greater than any acute rectilineal angle , and the remaining angle less than any rectilineal angle ...
Página 84
E B C First , let the segment BAED be greater than a semicircle . Take ( III . 1. ) F the centre of the circle ABCD , and join BF , FD . And because the angle BFD is at the centre , and the angle BAD at the circumference , and that they ...
E B C First , let the segment BAED be greater than a semicircle . Take ( III . 1. ) F the centre of the circle ABCD , and join BF , FD . And because the angle BFD is at the centre , and the angle BAD at the circumference , and that they ...
Página 85
For the same reason , because CBED is greater than a semicircle , 2 . The angles CAD , CED , are equal : therefore ( Ax . 2. ) 3 . The whole angle BAD is equal to the whole angle BED . Wherefore the angles in the same segment , & c .
For the same reason , because CBED is greater than a semicircle , 2 . The angles CAD , CED , are equal : therefore ( Ax . 2. ) 3 . The whole angle BAD is equal to the whole angle BED . Wherefore the angles in the same segment , & c .
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AB is equal ABC is equal ABCD angle ABC angle ACB angle AGH angle BAC angle BCD angle equal base BC bisected centre circle ABC circumference coincide common demonstrated describe diameter distance divided double draw equal angles exterior angle extremity fall figure four given circle given point given straight line gnomon greater impossible inscribed join less Let ABC Let the straight likewise manner meet opposite angles parallel parallelogram pass pentagon perpendicular point F PROBLEM produced Q.E.D. PROP reason rectangle contained rectilineal figure remaining angle right angles segment semicircle shown side BC sides square of AC straight line AC Take touches the circle triangle ABC twice the rectangle wherefore whole
Información bibliográfica
| Título | The Synoptical Euclid; Being the First Four Books of Euclid's Elements of Geometry from the Edition of Dr. Robert Simson ... With Exercises. By S. A. Good ... Second Edition |
| Autores | EUCLID., Samuel A. GOOD |
| Edición | 2 |
| Editor | Charles Henry Law, 1854 |
| Procedencia del original | Biblioteca Británica |
| Digitalizado | 10 Dic. 2013 |
| Largo | 120 páginas |
| Exportar cita | BiBTeX EndNote RefMan |