## 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

Resultat 1-5 av 5

Side 52

Book II . equal to the squares of EG , GF ; therefore the square of EF is

the square GF . and GF is equal b to CD ; therefore the h . 34. 1 . square of EF is

Book II . equal to the squares of EG , GF ; therefore the square of EF is

**double**ofthe square GF . and GF is equal b to CD ; therefore the h . 34. 1 . square of EF is

**double**of the square of CD . but the square of Е , AE is likewise**double**of the ... Side 53

3 . therefore the square of AG is

squares of AD , DG are equal i to the square of AG ; therefore the squares of AD ,

DG are

3 . therefore the square of AG is

**double**of the squares of AC , CD . but thesquares of AD , DG are equal i to the square of AG ; therefore the squares of AD ,

DG are

**double**of the squares of AC , CD . but DG is equal to DB ; therefore the ... Side 77

Let ABC be a circle , and BEC an angle at the center , and BAC Book III . an

angle at the circumference , which have the same circumference BC for their

base ; the angle BEC is

circle be ...

Let ABC be a circle , and BEC an angle at the center , and BAC Book III . an

angle at the circumference , which have the same circumference BC for their

base ; the angle BEC is

**double**of the angle BAC . A First , Let E the center of thecircle be ...

Side 105

KFC , and the angle BKF to FKC . wherefore the angle BFC is

KFC , and BKC

the angle CFL , and CLD

KFC , and the angle BKF to FKC . wherefore the angle BFC is

**double**of the angleKFC , and BKC

**double**of FKC . for the same reason , the angle CFD is**double**ofthe angle CFL , and CLD

**double**of CLF . and because the circumference BC is ... Side 121

Take any equimultiples of each of them , as the

5th of this Book , if the

, the

Take any equimultiples of each of them , as the

**doubles**of each . then by Def .5th of this Book , if the

**double**of the first be greater than the**double**of the second, the

**double**of the third is greater than the**double**of the fourth . but if the first be ...### 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.