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

### Inni boken

Resultat 1-5 av 100

Side 7

ET

ET

**A B**be a given finite**right line**: it is required to constitute an equilateral triangle upon the**right line**A B. From the centre A , with the distance**A B**, let the circle BCD be described [ by poftul . 3. ) and again from the centre ... Side 8

Let the two unequal

Let the two unequal

**right**lines given be**A B**and c , whereof let**A B**be the greater ; it is required to take a**right line**from**A B**the greater , equal to c the lefser . Put [ by prop ; 2. ] at the point a a**right line**A D , equal to the ... Side 9

of the

of the

**right**lines A E and c is equal to the**right line**Ad ; wherefore the**right line**A E is alio equal to the righi**line**c . Therefore two unequal**right**lires ,**A B**and c , being given , there is taken from the greater**A B**, a**line**... Side 10

... and the

... and the

**right line A B**upon the right line D E , then shall the point B agree with the point E ; because A B is equal to De : But since B CE the**right line A B**agrees with D E , the right line A c shall also A C shall also agree ... Side 11

continue the equal fides

continue the equal fides

**A B**, AC directly forwards to D and E : I say , the angłe ABC is equal to the angle ACB ; and the angle CBD equal to the angle BCE . For in the**right line**BD , let there be taken any point as F ; and from the ...### Hva folk mener - Skriv en omtale

Vi har ikke funnet noen omtaler på noen av de vanlige stedene.

### Andre utgaver - Vis alle

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...