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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Página 6
... shall at length meet upon that side on which are the angles which are less than two right angles . " PROPOSITION 1. - PROBLEM . To describe an equilateral triangle 6 EUCLID'S ELEMENTS .
... shall at length meet upon that side on which are the angles which are less than two right angles . " PROPOSITION 1. - PROBLEM . To describe an equilateral triangle 6 EUCLID'S ELEMENTS .
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To describe an equilateral triangle upon a given finite straight line . Let AB be the given straight line ; it is required to describe an equilateral triangle upon it . From the centre A , at the distance AB , describe ( Postulate 3. ) ...
To describe an equilateral triangle upon a given finite straight line . Let AB be the given straight line ; it is required to describe an equilateral triangle upon it . From the centre A , at the distance AB , describe ( Postulate 3. ) ...
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the straight line AD equal to C ; and from the centre A , and at the distance AD , describe ( Post . 3. ) the circle DEF : AE shall be equal to C. E B Because A is the centre of the circle DEF , ( Def . 15. ) 1 .
the straight line AD equal to C ; and from the centre A , and at the distance AD , describe ( Post . 3. ) the circle DEF : AE shall be equal to C. E B Because A is the centre of the circle DEF , ( Def . 15. ) 1 .
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off AE equal to AD ; join DE , and upon it , on the side remote from 4 , describe ( I. 1. ) an equilateral triangle DEF ; then join AF ; the straight line AF bisects the triangle BAC . A D F B Because AD is equal to AE , and AF is ...
off AE equal to AD ; join DE , and upon it , on the side remote from 4 , describe ( I. 1. ) an equilateral triangle DEF ; then join AF ; the straight line AF bisects the triangle BAC . A D F B Because AD is equal to AE , and AF is ...
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Describe ( I. 1. ) upon it an equilateral triangle ABC , and bisect ( I. 9. ) the angle ACB by the straight line CD . AB is cut into two equal parts in the point D. ) C A. B Because AC is equal to CB , and CD common to the two triangles ...
Describe ( I. 1. ) upon it an equilateral triangle ABC , and bisect ( I. 9. ) the angle ACB by the straight line CD . AB is cut into two equal parts in the point D. ) C A. B Because AC is equal to CB , and CD common to the two triangles ...
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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 |