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 |
Dentro del libro
Resultados 1-5 de 39
Página 10
to AG , and AB ( Hypothesis ) to AC , the two sides FA , AC are equal to the two GA , AB , each to each ; and they contain the angle FAG common to the two triangles AFC , ÁGB ; therefore ( I. 4. ) 1 .
to AG , and AB ( Hypothesis ) to AC , the two sides FA , AC are equal to the two GA , AB , each to each ; and they contain the angle FAG common to the two triangles AFC , ÁGB ; therefore ( I. 4. ) 1 .
Página 11
... one of them is greater than the other . Let AB be the greater , and from it cut off ( I. 3. ) DB equal to AC , the less , and join DC . Therefore because in the triangles DBC , ACB , DB is equal to AC , and BC common to both , 1 .
... one of them is greater than the other . Let AB be the greater , and from it cut off ( I. 3. ) DB equal to AC , the less , and join DC . Therefore because in the triangles DBC , ACB , DB is equal to AC , and BC common to both , 1 .
Página 13
A D F B Because AD is equal to AE , and AF is common to the two triangles DAF , EAF ; 1 . The troo sides DA , AF , are equal to the two sides E4 , AF , each to each ; and ( Constr . ) 2 .
A D F B Because AD is equal to AE , and AF is common to the two triangles DAF , EAF ; 1 . The troo sides DA , AF , are equal to the two sides E4 , AF , each to each ; and ( Constr . ) 2 .
Página 14
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 . The two sides AC , CD are equal to BC , CD , each to each ; and ( Constr . ) 2 .
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 . The two sides AC , CD are equal to BC , CD , each to each ; and ( Constr . ) 2 .
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 ...
Comentarios de la gente - Escribir un comentario
No encontramos ningún comentario en los lugares habituales.
Otras ediciones - Ver todas
Términos y frases comunes
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 |