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

### Inni boken

Resultat 1-5 av 71

Side 41

Let the parallelograms

Let the parallelograms

**ABCD**, EBCF be constituted upon the same base Bc , and between the same parallels AF , BC : I say , the parallelogram**ABCD**is equal to the parallelogram E BC F. For because**A B C D**is a parallelogram [ by prop . Side 42

rallels A H , BG : I say , the parallelograms

rallels A H , BG : I say , the parallelograms

**A B C D**, EF G H are equal to one another . For join B E , CH . Then because Bc is equal to FG , А D EL H and F G equal to EH ; therefore will bc be equal to E H. They are parallel too ... Side 45

... parallelogram ABC ) , and the triangle E B C have the same bale Bc , and be between the same parallels B C , AE : I say , the parallelogram

... parallelogram ABC ) , and the triangle E B C have the same bale Bc , and be between the same parallels B C , AE : I say , the parallelogram

**ABCD**will be double to the triangle B E C. For draw A C. Then the triangle A B C [ hy prop . Side 48

Let the given right - lined figure be

Let the given right - lined figure be

**ABCD**, and the given right - lined angle be E : It is required to constitute a parallelogram equal to the given right - lined figure AB CD , having one angle equal to the angle E. Draw D B ... Side 49

Therefore the parallelogram KFL M is constituted equal to the given right - lined figure

Therefore the parallelogram KFL M is constituted equal to the given right - lined figure

**A B C D**, having one angle FKM , which is equal to the given angle E. Which was to be done , PROP . XLVI . PROBL . To describe a Square upon a ...### 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...