The first three books of Euclid's Elements of geometry, with theorems and problems, by T. Tate |
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Resultat 1-5 av 76
Side 6
therefore ca is equal to CB ; wherefore ca , AB , BC are equal to one another ; and the triangle ABC is therefore equilateral , and it is described upon the given straight line A B. Which was required to be done . K D B L G E FU PROP .
therefore ca is equal to CB ; wherefore ca , AB , BC are equal to one another ; and the triangle ABC is therefore equilateral , and it is described upon the given straight line A B. Which was required to be done . K D B L G E FU PROP .
Side 7
But it has been shown that BC is equal to BG ; wherefore Al and BC are each of them equal to BG ; and things that are equal to the same are equal to one another ; therefore the straight line A L is equal to BC . Wherefore from the given ...
But it has been shown that BC is equal to BG ; wherefore Al and BC are each of them equal to BG ; and things that are equal to the same are equal to one another ; therefore the straight line A L is equal to BC . Wherefore from the given ...
Side 8
whence AE and c are each of them equal to AD ; wherefore the straight line AE is equal to c ( Ax . 1. ) , and from AB , the greater of two straight lines , a part A E has been cut off equal to c the less . Which was to be done . PROP .
whence AE and c are each of them equal to AD ; wherefore the straight line AE is equal to c ( Ax . 1. ) , and from AB , the greater of two straight lines , a part A E has been cut off equal to c the less . Which was to be done . PROP .
Side 9
wherefore also the point c shall coincide with the point F , because the straight line AC is equal to DF ( Hyp . ) . But the point B coincides with the point E ; wherefore the base BC shall coincide with the base EF , because the point ...
wherefore also the point c shall coincide with the point F , because the straight line AC is equal to DF ( Hyp . ) . But the point B coincides with the point E ; wherefore the base BC shall coincide with the base EF , because the point ...
Side 10
... therefore the two sides BF , Fc are equal to the two CG , G B , each to each ; and the angle byc is equal to the angle cGB , and the base BC is common to the two angles BFC , CGB ; wherefore the triangles are equal ( 1. 4. ) ...
... therefore the two sides BF , Fc are equal to the two CG , G B , each to each ; and the angle byc is equal to the angle cGB , and the base BC is common to the two angles BFC , CGB ; wherefore the triangles are equal ( 1. 4. ) ...
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Vanlige uttrykk og setninger
A B C ABCD adjacent angle ABC angle ACB angle BAC angle equal base BC BC is equal bisect centre circle ABC circumference coincide common construct demonstrated describe diameter divided double draw equal angles equal to FB equilateral exterior angle extremity figure fore four given point given straight line greater impossible interior join less Let ABC likewise lines AC manner meet opposite angles opposite sides parallel parallelogram pass perpendicular PROB produced Q. E. D. PROP rectangle A B rectangle contained remaining remaining angle right angles segment semicircle shown sides squares of AC straight line A B Take taken THEOR third touch touches the circle triangle ABC twice the rectangle Wherefore whole
Bibliografisk informasjon
| Tittel | The first three books of Euclid's Elements of geometry, with theorems and problems, by T. Tate Gleig's sch. series |
| Forfatter | Euclides |
| Redaktør | Thomas Tate |
| distribuert | 1851 |
| Original fra | Oxford University |
| Digitalisert | 30. jan 2009 |
| Lengde | 139 sider |
| Eksporter sitat | BiBTeX EndNote RefMan |