## The Elements of Euclid: Also the Book of Euclid's Data ...cor. viz. The first six books, together with the eleventh and twelfth |

### Inni boken

Side 368

tude CD ; CD together with a given magnitude has a

the ratio of AE to CD ) is given , as AE to CD , fo make BE to FD ; therefore the

ratio of BE to А E B FD is given , and BE is given , wherefore a . 2. Dat . FD is

given ...

tude CD ; CD together with a given magnitude has a

**given ratio**to AB . Becausethe ratio of AE to CD ) is given , as AE to CD , fo make BE to FD ; therefore the

ratio of BE to А E B FD is given , and BE is given , wherefore a . 2. Dat . FD is

given ...

Side 369

Let the excess of the magnitude AB above a given magnitude , have a

to the magnitude BC ; the excess of AC , both of them together , above a given

magnitude , has a

of ...

Let the excess of the magnitude AB above a given magnitude , have a

**given ratio**to the magnitude BC ; the excess of AC , both of them together , above a given

magnitude , has a

**given ratio**to BC . Let AD be the given magnitude the excessof ...

Side 370

Let AD be the given magnitude ; and because DB , the excess of AB above AD ,

has a

ratio of AD to DE the same with this ratio ; therefore the ratio of AD to DE is given

...

Let AD be the given magnitude ; and because DB , the excess of AB above AD ,

has a

**given ratio**to BC ; the ratio of DC to DB is a . 7. Dat . given a . make theratio of AD to DE the same with this ratio ; therefore the ratio of AD to DE is given

...

Side 375

5 . fore 6 AG is

than the

than AG . and AB , C AG are

5 . fore 6 AG is

**given**; and because the**ratio**of AB to CD is greater b . 2. Dat .than the

**ratio**of ( AE to CF , that A E is , than the**ratio**of ) AG to CD ; AB is greaterthan AG . and AB , C AG are

**given**, therefore the remainder BG is**given**. and ... Side 377

Let AB , CD , EF be three magnitudes , and let GD the excess of one of them CD

above the given magnitude CG have a

excess of the fame CD above the given magnitude CK have a

Let AB , CD , EF be three magnitudes , and let GD the excess of one of them CD

above the given magnitude CG have a

**given ratio**to AB ; and also let KD theexcess of the fame CD above the given magnitude CK have a

**given ratio**to EF ...### Hva folk mener - Skriv en omtale

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### Vanlige uttrykk og setninger

added alſo altitude angle ABC angle BAC baſe becauſe Book Book XI caſe circle circle ABCD circumference common cone cylinder Definition demonſtrated deſcribed diameter divided double draw drawn equal equal angles equiangular equimultiples exceſs fame fides firſt folid fore four fourth given angle given in poſition given in ſpecies given magnitude given ratio given ſtraight line greater Greek half join leſs likewiſe magnitude manner meet muſt oppoſite parallel parallelogram perpendicular plane priſm produced PROP proportionals Propoſition pyramid radius reaſon rectangle rectangle contained remaining right angles ſame ſecond ſegment ſhall ſhewn ſides ſimilar ſolid ſphere ſquare ſquare of AC taken THEOR theſe third thro triangle ABC wherefore whole

### Populære avsnitt

Side 157 - D ; wherefore the remaining angle at C is equal to the remaining angle at F ; Therefore the triangle ABC is equiangular to the triangle DEF. Next, let each of the angles at C, F be not less than a right angle ; the triangle ABC is also, in this case, equiangular to the triangle DEF.

Side 24 - ... be equal, each to each ; and also the third angle of the one to the third angle of the other. Let ABC, DEF be two triangles which have the angles ABC, ECA equal to the angles DEF, EFD, viz.

Side 45 - If a straight line be divided into any two parts, the rectangle contained by the whole and one of the parts, is equal to the rectangle contained by the two parts, together with the square of the aforesaid part.

Side 73 - THE straight line drawn at right angles to the diameter of a circle, from the extremity of...

Side 163 - 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 81 - ELF. Join BC, EF ; and because the circles ABC, DEF are equal, the straight lines drawn from their centres are equal: therefore the two sides BG, GC, are equal to the two EH, HF ; and the angle at G is equal to the angle at H ; therefore the base BC is equal (4.

Side 323 - Equiangular parallelograms have to one another the ratio which is compounded of the ratios of their sides.

Side 80 - EA : and because AD is equal to DC, and DE common to the triangles ADE, CDE, the two sides AD, DE are equal to the two CD, DE, each to each ; and the angle ADE is equal to the angle CDE, for each of them is a right angle ; therefore the base AE is equal (4.

Side 36 - To a given straight line to apply a parallelogram, which shall be equal to a given triangle, and have one of its angles equal to a given rectilineal angle.