## 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 8

Post . tance AD describe the circle DEF . and because A is the center of the circle

DEF , AE shall be equal to AD . but the straight line C is

whence AE and C are each of C. 1. Ax . them equal to AD . wherefore the straight

...

Post . tance AD describe the circle DEF . and because A is the center of the circle

DEF , AE shall be equal to AD . but the straight line C is

**likewise**equal to AD ,whence AE and C are each of C. 1. Ax . them equal to AD . wherefore the straight

...

Side 9

Therefore if two triangles hảve two sides of the one equal to two sides of the other

, each to each , and have

one another ; their bases shall

Therefore if two triangles hảve two sides of the one equal to two sides of the other

, each to each , and have

**likewise**the angles contained by those fides equal toone another ; their bases shall

**likewise**be equal , and the triangles be equal ... Side 13

I. the sides BA , AC cannot but coincide with the sides ED , DF ; wherefore

therefore if two triangles , & c . Q. E. D. b . 8. A. ТО b . I. I. PROP . IX . PRO B. 10

bisect a given ...

I. the sides BA , AC cannot but coincide with the sides ED , DF ; wherefore

**likewise**the angle BAC coincides with the angle EDF , and is equal to it .therefore if two triangles , & c . Q. E. D. b . 8. A. ТО b . I. I. PROP . IX . PRO B. 10

bisect a given ...

Side 124

Therefore b as A is to C , so is B to C.

has to B. for , having made the same conD A ftru & tion , D may in like manner be

shewn E B equal to E. therefore if F be greater than D , CF it is

Therefore b as A is to C , so is B to C.

**Likewise**C has the same ratio to A that ithas to B. for , having made the same conD A ftru & tion , D may in like manner be

shewn E B equal to E. therefore if F be greater than D , CF it is

**likewise**greater ... Side 134

op Book V. that KH , NM are equimultiples

of BE be greater than KH which is a multiple of the fame BE , NP

multiple of DF shall be greater than NM the multiple of the same DF ; and if KO be

...

op Book V. that KH , NM are equimultiples

**likewise**of BE , DF , if KO the multipleof BE be greater than KH which is a multiple of the fame BE , NP

**likewise**themultiple of DF shall be greater than NM the multiple of the same DF ; and if KO be

...

### 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.