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

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Resultat 1-5 av 7

Side 269

6.6 . and because it has been

AK is equal to KQ , and EM to MS , as OK to KL , so is SM to MN ; and therefore ,

the sides about the right angles OKL , SMN being proportionals , the triangle LKQ

...

6.6 . and because it has been

**shewn**that as AK to KL , so is EM to MN , and thatAK is equal to KQ , and EM to MS , as OK to KL , so is SM to MN ; and therefore ,

the sides about the right angles OKL , SMN being proportionals , the triangle LKQ

...

Side 282

... because in the preceding Proposition it was

circle FGH . and AZ is perpendicular to the plane KBOS , and is therefore the

shortest of all the straight lines that can be drawn from A the center of the sphere

to that ...

... because in the preceding Proposition it was

**shewn**that KV falls without thecircle FGH . and AZ is perpendicular to the plane KBOS , and is therefore the

shortest of all the straight lines that can be drawn from A the center of the sphere

to that ...

Side 285

... to BC ; which was

to any sphere greater than DEF the triplicate ratio of that which BC has to EF .

and it was demonstrated that neither has it that ratio to any sphere less than DEF

.

... to BC ; which was

**shewn**to be impossible . therefore the sphere ABC has notto any sphere greater than DEF the triplicate ratio of that which BC has to EF .

and it was demonstrated that neither has it that ratio to any sphere less than DEF

.

Side 350

The Demonstration of this is imperfect , because it is not

pyramids into which those upon multangular bafes are divided , are similar to one

another , as ought necessarily to have been done , and is done in the like ...

The Demonstration of this is imperfect , because it is not

**shewn**that the triangularpyramids into which those upon multangular bafes are divided , are similar to one

another , as ought necessarily to have been done , and is done in the like ...

Side 423

... as has been

can be described in the segment EGF upon the base EF , and the ratio of its

perpendicular to the base is the fame , as was there

to BC ...

... as has been

**shewn**in the preceding demonstration , a triangle similar to ABCcan be described in the segment EGF upon the base EF , and the ratio of its

perpendicular to the base is the fame , as was there

**shewn**, with the ratio of ADto BC ...

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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.