## Euclid's Elements of Geometry: The First Six, the Eleventh and Twelfth Books |

### Inni boken

Resultat 1-5 av 47

Side 55

... four fides are greater than

... four fides are greater than

**twice**the diagonals . PROP . III . THEOR . All the angles of any right - lined figure , when the angles are all inward , are equal to**twice**as many right angles , wanting four , as that figure bas fides ... Side 56

as many right angles as there are triangles ; that is , equal to

as many right angles as there are triangles ; that is , equal to

**twice**as many right angles , wanting the angles of two triangles ; that is , wanting four right angles , as the figure has fides . Therefore all the angles of any right ... Side 57

H equal to

H equal to

**twice**as many right angles as the figure has angles , that is , has fides . But since all the inward angles together , with four right angles , has been proved to be equal to**twice**as many right angles as the figure has sides ... Side 58

... added to 4 , and lefsened by

... added to 4 , and lefsened by

**twice**the number of sides of the figure ; as , if the figure has four fides , it can have one outward angle . And in this case , the outward angle is equal to all the three inward angles . Side 63

And if the angle B F C be added in common , the angles F A C , A F B , and

And if the angle B F C be added in common , the angles F A C , A F B , and

**twice**the angle B F C , will be equal to the angles BFC , CFD , CDF ; that is , to the angles BFD , CDF . And if the equal angles A F B , B F D be taken away ...### Hva folk mener - Skriv en omtale

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Euclid's Elements of Geometry: The First Six, the Eleventh and Twelfth Books Euclid,David Gregory Ingen forhåndsvisning tilgjengelig - 2016 |

### 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...