## 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 62

Wherefore if any two points , & c . Q. E. D. PROP . III . THEOR . 2. I. 3 • b . 8. I. [ F a

straight line drawn

does not pass

...

Wherefore if any two points , & c . Q. E. D. PROP . III . THEOR . 2. I. 3 • b . 8. I. [ F a

straight line drawn

**thro**' the center of a circle , bifect a straight line in it whichdoes not pass

**thro**' the center , it shall cut it at right angles . and if it cuts it at right...

Side 63

IF F in a circle two straight lines cut one another which do not both pass

center , they do not bisect each the other . a . 1. 3 . Let ABCD be a circle , and AC

, BD two straight lines in it which cut one another in the point E , and do not both ...

IF F in a circle two straight lines cut one another which do not both pass

**thro**' thecenter , they do not bisect each the other . a . 1. 3 . Let ABCD be a circle , and AC

, BD two straight lines in it which cut one another in the point E , and do not both ...

Side 66

... circumference equal to FG . for if there can , let it be FK , and because FK is

equal to FG , and FG to FH , FK is equal to FH , that is , a line nearer to that which

passes

impossible .

... circumference equal to FG . for if there can , let it be FK , and because FK is

equal to FG , and FG to FH , FK is equal to FH , that is , a line nearer to that which

passes

**thro**' the center is equal to one which is more remote ; which isimpossible .

Side 89

But let one of them BD pass

not pass

in F , F is the center of the circle ABCD ; and join AF . and because BD ...

But let one of them BD pass

**thro**' the center , and cut the other AC , which doesnot pass

**thro**' the center , at right angles , in the point E. then , if BD be bisectedin F , F is the center of the circle ABCD ; and join AF . and because BD ...

Side 91

Either DCA pafles

E , and join EB ; D therefore the angle EBD is a right a angle . and because the

straight line a . 18. 3 . AC is bisected in E , and produced to the point D , the ...

Either DCA pafles

**thro**' the center , or it does not ; first , let it pass**thro**' the centerE , and join EB ; D therefore the angle EBD is a right a angle . and because the

straight line a . 18. 3 . AC is bisected in E , and produced to the point D , the ...

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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.