## 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 66

... the two sides GE , EF are equal to the two HE , EF ; and the angle E GEFis

equal to the angle HEF , therefore the base FG is equal d to the base FH . but

besides FH no other straight line can be drawn from F to D H н the

equal ...

... the two sides GE , EF are equal to the two HE , EF ; and the angle E GEFis

equal to the angle HEF , therefore the base FG is equal d to the base FH . but

besides FH no other straight line can be drawn from F to D H н the

**circumference**equal ...

Side 68

F a point be taken within a circle , from which there fall more than two equal

straight lines to the

point D be taken within the circle ABC , from which to the

more ...

F a point be taken within a circle , from which there fall more than two equal

straight lines to the

**circumference**, that point is the center of the circle . Let thepoint D be taken within the circle ABC , from which to the

**circumference**there fallmore ...

Side 184

Book VI . the

FHF . and if the

the angle BGL is also equal a to the angle EHN ; and if the

Book VI . the

**circumference**EF , the same multiple is the angle EHN of the angleFHF . and if the

**circumference**BL be equal to the circumfe . & . 37. 3 . rence EN ,the angle BGL is also equal a to the angle EHN ; and if the

**circumference**BL be ... Side 185

... MHN may be proved equal to one another . therefore what multiple soever the

of the sector BGC . for the same reason , whatever multiple the

... MHN may be proved equal to one another . therefore what multiple soever the

**circumference**BL is of the**circumference**BC , the same multiple is the sector BGLof the sector BGC . for the same reason , whatever multiple the

**circumference**... Side 1

Elem . the angle ABC is to a right angle ABD , as the arch AC to the arch AD ; and

quadrupling the consequents , the angle ABC will be to four right angles , as the

arch AC to four times the arch AD , or to the whole

Elem . the angle ABC is to a right angle ABD , as the arch AC to the arch AD ; and

quadrupling the consequents , the angle ABC will be to four right angles , as the

arch AC to four times the arch AD , or to the whole

**circumference**. LEMMA II .### 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.