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 |
Inni boken
Side 113
... Excess of the first above the second , is to the second , as the Excefs of the third above the fourth , is to the fourth . 17th Prop . Book 5th . XVII . Convertendo , by Converfion ; when there are four proportionals , and it is ...
... Excess of the first above the second , is to the second , as the Excefs of the third above the fourth , is to the fourth . 17th Prop . Book 5th . XVII . Convertendo , by Converfion ; when there are four proportionals , and it is ...
Side 136
... excess above the fourth , Let AB be to BE , as CD to DF ; then BA is to AE , as DC to CF. Because AB is to BE , as CD to DF , by E divifion 2 , AE is to EB , as CF to FD ; and by inverfion , BE is to EA , as DF to FC . Wherefore , by ...
... excess above the fourth , Let AB be to BE , as CD to DF ; then BA is to AE , as DC to CF. Because AB is to BE , as CD to DF , by E divifion 2 , AE is to EB , as CF to FD ; and by inverfion , BE is to EA , as DF to FC . Wherefore , by ...
Side 141
... excess of the first and fifth fhall be to the fecond , as the excess of the third and fixth to the fourth , the Demonftration of this is the Book V. fame with that of the Propofition , if OF EUCLID . 14.
... excess of the first and fifth fhall be to the fecond , as the excess of the third and fixth to the fourth , the Demonftration of this is the Book V. fame with that of the Propofition , if OF EUCLID . 14.
Side 178
... than C. Make the parallelogram KLMN equal to the excess of EF above C , and fimilar and fimi- d . 21. 6. larly fituated to D ; but D is fimilar to EF , therefore alfo KM is fimilar to EF . let KL be the homologous fide 178 THE ELEMENTS.
... than C. Make the parallelogram KLMN equal to the excess of EF above C , and fimilar and fimi- d . 21. 6. larly fituated to D ; but D is fimilar to EF , therefore alfo KM is fimilar to EF . let KL be the homologous fide 178 THE ELEMENTS.
Side 212
... excess of the square of AB above the square of LX , and make RX equal to its fide , and join RL , RM , RN . because RX is perpendicular to the plane of the m . 3. Def . circle LMN , it is m perpendicular to each of the ftraight lines LX ...
... excess of the square of AB above the square of LX , and make RX equal to its fide , and join RL , RM , RN . because RX is perpendicular to the plane of the m . 3. Def . circle LMN , it is m perpendicular to each of the ftraight lines LX ...
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.