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 7
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 ... D B C 9 Because the point A is the centre of the circle BCD , ( Definition 15. ) 1 .
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 ... D B C 9 Because the point A is the centre of the circle BCD , ( Definition 15. ) 1 .
Página 8
Wherefore from the given point A a straight line AL has been drawn equal to the given straight line BC . Which was to be done . PROP . III . - PROBLEM , From the greater of two given straight lines to cut off a part equal to the less .
Wherefore from the given point A a straight line AL has been drawn equal to the given straight line BC . Which was to be done . PROP . III . - PROBLEM , From the greater of two given straight lines to cut off a part equal to the less .
Página 13
To bisect a given rectilineal angle , that is , to divide it into troo equal angles . Let BAC be the given rectilineal angle , it is required to bisect it . Take any point D in AB , and from AC cut ( I. 3. ) off AE equal to AD ; join DE ...
To bisect a given rectilineal angle , that is , to divide it into troo equal angles . Let BAC be the given rectilineal angle , it is required to bisect it . Take any point D in AB , and from AC cut ( I. 3. ) off AE equal to AD ; join DE ...
Página 14
To bisect a given finite straight line , that is , to divide it into two equal parts . ... 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 ACD , BCD ; 1 .
To bisect a given finite straight line , that is , to divide it into two equal parts . ... 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 ACD , BCD ; 1 .
Página 15
straight line FC drawn from the given point C is at right angles to the given straight line AB E B A D Because DC is equal to CE , and FC common to the two triangles DCF , ECF ; 1 . The two sides DC , CF , are equal to the two EC ...
straight line FC drawn from the given point C is at right angles to the given straight line AB E B A D Because DC is equal to CE , and FC common to the two triangles DCF , ECF ; 1 . The two sides DC , CF , are equal to the two EC ...
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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 |