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
From the point A to B draw ( Post . I. ) the straight line AB ; and upon it describe ( I. 1. ) the equilateral_triangle DAB , and produce ( Post . 2. ) the straight lines DĂ , DB to E and F. From the centre B at the distance BC ...
From the point A to B draw ( Post . I. ) the straight line AB ; and upon it describe ( I. 1. ) the equilateral_triangle DAB , and produce ( Post . 2. ) the straight lines DĂ , DB to E and F. From the centre B at the distance BC ...
Página 9
The point C shall coincide with the point F , because the straight line AC is equal to DF . But the point B coincides with the point E ; wherefore the base BC shall coincide with the base EF ; because , the point B coinciding with E ...
The point C shall coincide with the point F , because the straight line AC is equal to DF . But the point B coincides with the point E ; wherefore the base BC shall coincide with the base EF ; because , the point B coinciding with E ...
Página 10
A B 1 G D E و In BD take any point F , and from AE , the greater , cut off AG equal ( I. 3. ) to AF , the less , and join FC , GB . Because ÅF is equal ( Constr . ) to AG , and AB ( Hypothesis ) to AC , the two sides FA , AC are equal ...
A B 1 G D E و In BD take any point F , and from AE , the greater , cut off AG equal ( I. 3. ) to AF , the less , and join FC , GB . Because ÅF is equal ( Constr . ) to AG , and AB ( Hypothesis ) to AC , the two sides FA , AC are equal ...
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D G А F B For if the triangle ABC be applied to DEF , so that the point B be on E , and the straight line BC upon EF ; 1 . The point C shall coincide with the point F , because BC ' is equal to EF ; therefore BC coinciding with EF , 2 .
D G А F B For if the triangle ABC be applied to DEF , so that the point B be on E , and the straight line BC upon EF ; 1 . The point C shall coincide with the point F , because BC ' is equal to EF ; therefore BC coinciding with EF , 2 .
Página 16
the circle EGF meeting , AB in F , G ; and bisect ( I. 10. ) FG in H , and join CH ; the straight line CH , drawn from the given point C , is perpendicular to the given straight line AB . E HI A F G B Join CF , CG ; and because FH is ...
the circle EGF meeting , AB in F , G ; and bisect ( I. 10. ) FG in H , and join CH ; the straight line CH , drawn from the given point C , is perpendicular to the given straight line AB . E HI A F G B Join CF , CG ; and because FH is ...
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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 |