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

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Side 10

Which was to be demonstrated . w Some , for want of making a

Which was to be demonstrated . w Some , for want of making a

**difference**between a geometrical congruency and a mechanical one ; that is , between an intellectual or mental one , and an actual sensual one made with the hands and the eyes ... Side 60

... will be equal to two right angles ; and the

... will be equal to two right angles ; and the

**difference**of the angles contained under the equal sides , will be equal to the**difference**of the said acute and oblique angles , Let 1 eLet Ą B C , Der be two triangles , 60 ADDITIONS . Side 61

... and the

... and the

**difference**of the angles A B C , DE F , contained by the equal fides A B , BC , and DE , EF , will be equal to the**difference**of the obtuse and acute angles A CE , DFE , opposite to the equal fides A B , DE . Side 62

Consequently , the

Consequently , the

**difference**of the angles D E F and A B C , will be equal to the**difference**of the angles A CB , EFD . Therefore if two triangles have two sides of the one equal to two sides of the other , each to each ; and one angle ... Side 63

Therefore the angle cd F exceeds the angle Faç by twice the angle BFC ; and so the angle byc is one half their

Therefore the angle cd F exceeds the angle Faç by twice the angle BFC ; and so the angle byc is one half their

**difference**. Secondly , Let the perpendicular F c fall without the triangle A FD . Then because the external angle A D F of ...### 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 cylinder definition demonſtrated deſcribed diameter difference divided double draw drawn equal equal angles equiangular equimultiples Euclid exceeds fall fame fides figure firſt folid fore 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 247 - 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 30 - 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 248 - 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 18 - 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 32 - 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 56 - 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 391 - 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 110 - If any two points be taken in the circumference of a circle, the straight line which joins them shall fall within the circle.

Side 130 - When you have proved that the three angles of every triangle are equal to two right angles...

Side 183 - 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...