The Elements of Euclid: Also the Book of Euclid's Data ...cor. viz. The first six books, together with the eleventh and twelfthF. Wingrave, 1804 |
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Side 38
... wherefore the angles BHF , HFE are lefs than two right angles . but straight lines which with another straight line make the interior angles upon the d . 12. Ax . fame fide lefs than two right angles , do meet if produced far enough ...
... wherefore the angles BHF , HFE are lefs than two right angles . but straight lines which with another straight line make the interior angles upon the d . 12. Ax . fame fide lefs than two right angles , do meet if produced far enough ...
Side 46
... wherefore the angle CGB is equal to the angle GBC , and therefore the fide BC is equal to the fide CG . but CB is equal alfo f to GK , and CG to BK ; wherefore the figure CGKB is equilateral . it is like- wife rectangular ; for CG is ...
... wherefore the angle CGB is equal to the angle GBC , and therefore the fide BC is equal to the fide CG . but CB is equal alfo f to GK , and CG to BK ; wherefore the figure CGKB is equilateral . it is like- wife rectangular ; for CG is ...
Side 50
... wherefore AG alfo is equal to RF . therefore the four rectangles AG , MP , PL , RF are equal to one another , and so are quadruple of one of them AG . And it was demonstrated that the four CK , BN , GR , RN are quadruple of CK ...
... wherefore AG alfo is equal to RF . therefore the four rectangles AG , MP , PL , RF are equal to one another , and so are quadruple of one of them AG . And it was demonstrated that the four CK , BN , GR , RN are quadruple of CK ...
Side 53
... wherefore alfo the fide BD is equal to the fide DG . g . 6. I. again , because EGF is half a right angle , and that the an- gle at F is a right angle , be- cause it is equal to the op- pofite angle ECD , the re- maining angle FEG is ...
... wherefore alfo the fide BD is equal to the fide DG . g . 6. I. again , because EGF is half a right angle , and that the an- gle at F is a right angle , be- cause it is equal to the op- pofite angle ECD , the re- maining angle FEG is ...
Side 62
... wherefore DF is greater than DE , the less than the greater , which is impoffible . therefore the straight line drawn from A to B does not fall with- out the circle . in the fame manner , it may be demonstrated that it does not fall ...
... wherefore DF is greater than DE , the less than the greater , which is impoffible . therefore the straight line drawn from A to B does not fall with- out the circle . in the fame manner , it may be demonstrated that it does not fall ...
Vanlige uttrykk og setninger
alfo alſo angle ABC angle BAC bafe baſe BC is equal BC is given becauſe the angle becauſe the ratio bifected Book XI cafe circle ABCD circumference cone conſtruction cylinder defcribed demonſtrated drawn EFGH equal angles equiangular equimultiples Euclid excefs faid fame multiple fame ratio fecond fegment fhall fhewn fides fides BA fimilar fince firft firſt folid angle fore fquare fquare of AC given angle given in fpecies given in magnitude given in pofition given magnitude given ratio given ſtraight line gnomon greater join lefs leſs likewife line BC muſt oppofite parallel parallelepipeds parallelogram perpendicular plane angles PROP Propofition pyramid rectangle contained rectilineal figure right angles ſame ſhall ſpace ſphere ſquare ſtraight line AB THEOR theſe thro tiple triangle ABC wherefore
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.