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 34
The whole angle ABD is equal to the whole angle ACD ; and the angle BAC has been shown to be equal to the angle BDC ; therefore the opposite sides and angles of parallelograms are equal to one another . Also , their diameter bisects ...
The whole angle ABD is equal to the whole angle ACD ; and the angle BAC has been shown to be equal to the angle BDC ; therefore the opposite sides and angles of parallelograms are equal to one another . Also , their diameter bisects ...
Página 35
The whole , or the remainder , AE is equal to the whole , or the remainder , DF ; AB also is equal ( I. 34. ) to DC ; and the two EA , AB , are therefore equal to the two FD , DC , each to each ; and the exterior angle FDC is equal ( I.
The whole , or the remainder , AE is equal to the whole , or the remainder , DF ; AB also is equal ( I. 34. ) to DC ; and the two EA , AB , are therefore equal to the two FD , DC , each to each ; and the exterior angle FDC is equal ( I.
Página 40
Let ABCD be a parallelogram , of which the diameter is AC , and EH , FG , the parallelograms about AC , that is , through which AC passes , and BK , KD , the other parallelograms which make up the whole figure ABCD , which are therefore ...
Let ABCD be a parallelogram , of which the diameter is AC , and EH , FG , the parallelograms about AC , that is , through which AC passes , and BK , KD , the other parallelograms which make up the whole figure ABCD , which are therefore ...
Página 43
The whole rectilineal figure ABCD is equal to the whole parallelogram KFLM . Therefore the parallelogram KFLM has been described equal to the given rectilineal figure ABCD , having the angle FKM equal to the given angle E. Which was to ...
The whole rectilineal figure ABCD is equal to the whole parallelogram KFLM . Therefore the parallelogram KFLM has been described equal to the given rectilineal figure ABCD , having the angle FKM equal to the given angle E. Which was to ...
Página 44
AB and AH are in the same straight line . And because the angle DBC is equal to the angle FB4 , each of them being a right angle , add to each the angle ABC , and ( Ax . 2. ) 3 . The whole angle DBA is equal to the whole FBC .
AB and AH are in the same straight line . And because the angle DBC is equal to the angle FB4 , each of them being a right angle , add to each the angle ABC , and ( Ax . 2. ) 3 . The whole angle DBA is equal to the whole FBC .
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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 |