## 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 6

Side 99

O inscribe a square in a given

required to inscribe a square in

angles to one another ; and join AB , BC , CD , DA . because BE is equal to ED ,

for E is the ...

O inscribe a square in a given

**circle**. a Let**ABCD**be the given**circle**; it isrequired to inscribe a square in

**ABCD**. Draw the diameters AC , BD at rightangles to one another ; and join AB , BC , CD , DA . because BE is equal to ED ,

for E is the ...

Side 246

THEOR . See N. Cdiameters MIRCLES are to one another as the squares of their

diameters . Let ABCD , EFGH be two circles , and BD , FH their diameters . as the

square of BD to the square of FH , so is the

THEOR . See N. Cdiameters MIRCLES are to one another as the squares of their

diameters . Let ABCD , EFGH be two circles , and BD , FH their diameters . as the

square of BD to the square of FH , so is the

**circle ABCD**to the circle EFGH . Side 248

Book XII . is to the space S. therefore as the

the polygon AXBOCPDR to the polygon EKFLGMHN . but the

greater than the polygon contained in it ; whered . 14. 5. fore the space S is

greater ...

Book XII . is to the space S. therefore as the

**circle ABCD**is to the space S , lo is ethe polygon AXBOCPDR to the polygon EKFLGMHN . but the

**circle ABCD**isgreater than the polygon contained in it ; whered . 14. 5. fore the space S is

greater ...

Side 249

the square of BD , so is the circle EFGH to a space less than the Book XII .

BD is not to the square of FH , as the

...

the square of BD , so is the circle EFGH to a space less than the Book XII .

**circle****ABCD**, which has been demonstrated to be impossible . therefore the square ofBD is not to the square of FH , as the

**circle ABCD**is to any space greater than the...

Side 267

to any not to the circle EFGH , as the cone AL to any solid which is less Book XII .

than the cone EN . In the same manner it may be demonstrated that the circle

EFGH is not to the

to any not to the circle EFGH , as the cone AL to any solid which is less Book XII .

than the cone EN . In the same manner it may be demonstrated that the circle

EFGH is not to the

**circle ABCD**, as the cone EN solid less than the cone AL .### Hva folk mener - Skriv en omtale

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

### 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.