The Elements of Euclid: Also the Book of Euclid's Data ...cor. viz. The first six books, together with the eleventh and twelfth |
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Side 14
Book I. Because AC is equal to CB , and CD common to the two triangles ACD ,
BCD ; the two fides AC , CD are equal to BC , CD , each to each ; and the angle
ACD is equal to the angle BCD ; therefore the base AD is equal to the base o DB
...
Book I. Because AC is equal to CB , and CD common to the two triangles ACD ,
BCD ; the two fides AC , CD are equal to BC , CD , each to each ; and the angle
ACD is equal to the angle BCD ; therefore the base AD is equal to the base o DB
...
Side 204
Book XI . to their common section , are also at right angles to the other plane d ;
and any straight line FG in the plane DE , which is at 4.4.Def.11 . right angles to
CE the common section of the planes , has been proved to be perpendicular to
the ...
Book XI . to their common section , are also at right angles to the other plane d ;
and any straight line FG in the plane DE , which is at 4.4.Def.11 . right angles to
CE the common section of the planes , has been proved to be perpendicular to
the ...
Side 334
F fame manner the triangle EBC is similar to the triangle FBC , and the triangle
EAC to FAC . therefore there are two solid figures each of which is contained by
six triangles , one of them by three triangles the common vertex of which is the ...
F fame manner the triangle EBC is similar to the triangle FBC , and the triangle
EAC to FAC . therefore there are two solid figures each of which is contained by
six triangles , one of them by three triangles the common vertex of which is the ...
Side 340
Pouk XI . these points common to both . Now , as it has not been shewn in Euclid
, that they cannot have a common segment , this does not prove that they cannot
meet in two points , from which their not having a common segment is deduced ...
Pouk XI . these points common to both . Now , as it has not been shewn in Euclid
, that they cannot have a common segment , this does not prove that they cannot
meet in two points , from which their not having a common segment is deduced ...
Side 15
in the same manner the plane FHG is perpendicular to the plane ADB ; and
therefore GF the .common section of the planes FEG , FHG is perpendicular to
the plane ADB ; ( 19. II . ) and because the angle FHG is the inclination of the
planes ...
in the same manner the plane FHG is perpendicular to the plane ADB ; and
therefore GF the .common section of the planes FEG , FHG is perpendicular to
the plane ADB ; ( 19. II . ) and because the angle FHG is the inclination of the
planes ...
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Vanlige uttrykk og setninger
added alſo altitude angle ABC angle BAC baſe becauſe Book Book XI caſe circle circle ABCD circumference common cone cylinder Definition demonſtrated deſcribed diameter divided double draw drawn equal equal angles equiangular equimultiples exceſs fame fides firſt folid fore four fourth given angle given in poſition given in ſpecies given magnitude given ratio given ſtraight line greater Greek half join leſs likewiſe magnitude manner meet muſt oppoſite parallel parallelogram perpendicular plane priſm produced PROP proportionals Propoſition pyramid radius reaſon rectangle rectangle contained remaining right angles ſame ſecond ſegment ſhall ſhewn ſides ſimilar ſolid ſphere ſquare ſquare of AC taken THEOR theſe third thro triangle ABC wherefore whole
Populære avsnitt
Side 157 - D ; wherefore the remaining angle at C is equal to the remaining angle at F ; Therefore the triangle ABC is equiangular to the triangle DEF. Next, let each of the angles at C, F be not less than a right angle ; the triangle ABC is also, in this case, equiangular to the triangle DEF.
Side 24 - ... be equal, each to each ; and also the third angle of the one to the third angle of the other. Let ABC, DEF be two triangles which have the angles ABC, ECA equal to the angles DEF, EFD, viz.
Side 45 - If a straight line be divided into any two parts, the rectangle contained by the whole and one of the parts, is equal to the rectangle contained by the two parts, together with the square of the aforesaid part.
Side 73 - THE straight line drawn at right angles to the diameter of a circle, from the extremity of...
Side 163 - If two triangles have one angle of the one equal to one angle of the other and the sides about these equal angles proportional, the triangles are similar.
Side 81 - ELF. Join BC, EF ; and because the circles ABC, DEF are equal, the straight lines drawn from their centres are equal: therefore the two sides BG, GC, are equal to the two EH, HF ; and the angle at G is equal to the angle at H ; therefore the base BC is equal (4.
Side 323 - Equiangular parallelograms have to one another the ratio which is compounded of the ratios of their sides.
Side 80 - EA : and because AD is equal to DC, and DE common to the triangles ADE, CDE, the two sides AD, DE are equal to the two CD, DE, each to each ; and the angle ADE is equal to the angle CDE, for each of them is a right angle ; therefore the base AE is equal (4.
Side 36 - To a given straight line to apply a parallelogram, which shall be equal to a given triangle, and have one of its angles equal to a given rectilineal angle.
Bibliografisk informasjon
| Tittel | The Elements of Euclid: Also the Book of Euclid's Data ...cor. viz. The first six books, together with the eleventh and twelfth The Elements of Euclid: Also the Book of Euclid's Data ...cor. Viz. The First Six Books, Together with the Eleventh and Twelfth, Robert Simson |
| Forfatter | Robert Simson |
| Utgiver | F. Wingrave, 1804 |
| Original fra | University of Michigan |
| Digitalisert | 11. jun 2007 |
| Eksporter sitat | BiBTeX EndNote RefMan |