 | Matthew Iley - 1820 - 512 Seiten
...and parallel lines are best made with a parallel ruler. PROBLEM v. To bisect a given finite straight line, that is, to divide it into two equal parts. Let AB be a given finite straight line to be divided into two equal parts. y A From A as a centre, with any opening... | |
 | William Jillard Hort - 1822 - 308 Seiten
...triangle ABC, is equilateral ; and it is described upon the right line AB, as was required to be done. PROBLEM. To bisect a given finite right line ; that is, to divide it into two equal parts. AL Let AC be the given right line ; it is required to divide it into two equal parts. Upon that given... | |
 | Peter Nicholson - 1825 - 1046 Seiten
...line AF. Which was to be done. Proportion X. Problem. To bisect a given finite straight line, that it, to divide it into two equal parts. Let AB be the given straight line j it is required to divide it into two equal parts. Describe( 1 . 1 . ) upon it an equilateral... | |
 | Euclides - 1826 - 226 Seiten
...equal to the two EA, AE, each to each; and the base DF is equal to the base EF: therefore the angle 0 DAF is equal to the angle EAF. Wherefore the given...line, that is, to divide it into two equal parts. line ; it is required to bisect it. Upon it a describe the equilateral triangle ABC ; and bisect the... | |
 | Robert Simson - 1827 - 546 Seiten
...bisected by the straight line AF. Which was to be done. PROP. X. PROB. To bisect a given Jinite straight line, that is, to divide it into two equal parts. Let AB be the given straight line ; it is required to divide it into two equal parts. Describe* upon it an equilateral... | |
 | John Playfair - 1829 - 210 Seiten
...four, eight, sixteen, &c. equal parts. 12 1>. PROPOSITION X. PROBLEM. To bisect a given finite straight line; that is, to divide it into two equal parts. Let AB be the given straight line; it is required to divide it into two equal parts. On AB describe an equilateral triangle... | |
 | Euclides - 1834 - 518 Seiten
...each to each ; and the base DF is t Constr. equal t to the base EF ; therefore the angle • 8. 1. DAF is equal* to the angle EAF: wherefore, the given rectilineal angle BAC is bisected by the straight line AF. Which was to be done. • 1. I. * 9. l. f Constr. f Cumtr. PROPOSITION X. PROR. —... | |
 | Euclid - 1835 - 540 Seiten
...bisected by the straight line AF. Which was to be done. PROP. X. PROB. To bisect a given finite straight line, that is, to divide it into two equal parts. Let AB be the given straight line ; it is required to divide it into two equal parts. al. 1. Describe a upon it an equilateral... | |
 | Andrew Bell - 1837 - 290 Seiten
...BAC is bisected by the straight line AF. PROPOSITION X. PROBLEM. To bisect a given finite straight line — that is, to divide it into two equal parts. Let AB be the given straight line ; it is required to divide it into two equal parts. Describe (I. 1) upon it an equilateral... | |
 | Euclides - 1838 - 264 Seiten
...AF, each to each ; and the base ico.«t. DF, is equalf to the base EF ; therefore the angle »s "1 DAF is equal* to the angle EAF : wherefore the given rectilineal angle BAC is bisected by the straight line AF. Which was to be done. PROP. X. PROB. Tobiseet a given finite straight line, that... | |
| |
To bisect a given finite straight line, that is, to divide it into two equal parts. Let AB be the given straight line : it is required to divide it intotwo equal parts. 
