Euclid's Elements of Geometry: The First Six, the Eleventh and Twelfth Books |
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Resultat 1-5 av 51
Side 20
Then because the right line A B stands upon the right A1 line C B E , the angles A B C , A BE are [ by prop . 13. ] equal to two , right angles ; but [ by the suppofition ] the angles A B C , A B D are I also equal to two right angles ...
Then because the right line A B stands upon the right A1 line C B E , the angles A B C , A BE are [ by prop . 13. ] equal to two , right angles ; but [ by the suppofition ] the angles A B C , A B D are I also equal to two right angles ...
Side 21
For because the right line A E stands upon the right line CD , D making the angles C E A , A E D. E Therefore the angles CEA , AED are [ by prop . 13. ] equal B to two right angles . Again , because the right line de stands upon the ...
For because the right line A E stands upon the right line CD , D making the angles C E A , A E D. E Therefore the angles CEA , AED are [ by prop . 13. ] equal B to two right angles . Again , because the right line de stands upon the ...
Side 42
is equal to the parallelagram D BCF , for they stand both upon the same base á c , and are between the same pas C rallelş B C , EF . But the triangle E angle A B C is the one half of the 42 Euclid's Elements . Book I.
is equal to the parallelagram D BCF , for they stand both upon the same base á c , and are between the same pas C rallelş B C , EF . But the triangle E angle A B C is the one half of the 42 Euclid's Elements . Book I.
Side 43
... for they stand upon equal bases BC , EF , between the same parallels BF , GH . But the triangle ABC is one half of the parallelogram GBC A , because [ by prop . 34. ) the diameter A B cuts it in halves ; and the triangle F E D is ...
... for they stand upon equal bases BC , EF , between the same parallels BF , GH . But the triangle ABC is one half of the parallelogram GBC A , because [ by prop . 34. ) the diameter A B cuts it in halves ; and the triangle F E D is ...
Side 46
will be equal to the triangle A E c ; for they stand upon equal bases B E , E C , and are between the same parallels B C , AG . Therefore the triangle A B C is double to the triangle A E C. But [ by prop . 41. ] ...
will be equal to the triangle A E c ; for they stand upon equal bases B E , E C , and are between the same parallels B C , AG . Therefore the triangle A B C is double to the triangle A E C. But [ by prop . 41. ] ...
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Vanlige uttrykk og setninger
A B C ABCD added alſo altitude baſe becauſe centre circle circumference common cone continue cylinder definition demonſtrated deſcribed diameter difference divided double draw drawn equal equal angles equiangular equimultiples Euclid exceeds fall fame fides figure firſt folid fore four fourth given right line greater half inſcribed join leſs magnitudes manner meet multiple oppoſite parallel parallelogram perpendicular plane polygon priſms PROP proportional propoſition proved pyramid ratio rectangle remaining angle right angles right line A B right lined figure ſame ſay ſecond ſegment ſhall ſides ſimilar ſince ſolid ſome ſphere ſquare ſtand ſum taken THEOR theſe third thoſe thro touch triangle triangle ABC twice vertex Wherefore whole whoſe baſe
Populære avsnitt
Side 245 - 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 28 - If two triangles have two angles of the one equal to two angles of the other, each to each, and one side equal to one side, viz. either the sides adjacent to the equal...
Side 246 - But it was proved that the angle AGB is equal to the angle at F ; therefore the angle at F is greater than a right angle : But by the hypothesis, it is less than a right angle ; which is absurd.
Side 16 - When a straight line set up on a straight line makes the adjacent angles equal to one another, each of the equal angles is right, and the straight line standing on the other is called a perpendicular to that on which it stands.
Side 30 - Let the straight line EF, which falls upon the two straight lines AB, CD, make the alternate angles AEF, EFD equal to one another; AB is parallel to CD.
Side 54 - Therefore all the angles of the figure, together with four right angles, are equal to twice as many right angles as the figure has sides.
Side 389 - KL: but the cylinder CM is equal to the cylinder EB, and the axis LN to the axis GH; therefore as the cylinder EB to...
Side 108 - If any two points be taken in the circumference of a circle, the straight line which joins them shall fall within the circle.
Side 128 - When you have proved that the three angles of every triangle are equal to two right angles...
Side 181 - FK : in the same manner it may be demonstrated, that FL, FM, FG are each of them equal to FH, or FK : therefore the five straight lines FG, FH, FK, FL, FM are equal to one another : wherefore the circle described from the centre F, at the distance of...