## 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 5

Side 79

If it be possible , let the two similar

because the

11.

If it be possible , let the two similar

**segments**of circles , viz . ... join CA , DA . andbecause the

**segment**ACB is fimilar to the**segment**A B ADB , and that similar**segments**of circles contain equal angles ; the angle ACB is equal to the angle b.11.

Side 80

Book III . because AB is equal to CD . therefore the straight line AB coinciding

with CD , the

is equal to it . Wherefore similar sego ments , . & c . l . E. D. 2. 23. 3 . PROP . XXV .

Book III . because AB is equal to CD . therefore the straight line AB coinciding

with CD , the

**segment**AEB multe coincide with the**segment**CFD , and thereforeis equal to it . Wherefore similar sego ments , . & c . l . E. D. 2. 23. 3 . PROP . XXV .

Side 81

... greater than the angle BAD , the center E falls without the

therefore is less than a semicircle . but ... greater than a semicircle . wherefore a

... greater than the angle BAD , the center E falls without the

**segment**ABC , whichtherefore is less than a semicircle . but ... greater than a semicircle . wherefore a

**segment**of a circle being given , the circle is described of which it is a**segment**. Side 85

IN D N a circle , the angle in a semicircle is a right angle ; but the angle in a

the diameter is BC , and center E ; and draw CA dividing the circle into the

IN D N a circle , the angle in a semicircle is a right angle ; but the angle in a

**segment**greater than a semicircle is less than a ... Let ABCD be a circle , of whichthe diameter is BC , and center E ; and draw CA dividing the circle into the

**segments**... Side 449

PROP . XCVIII . in jier , P. 2 IF a straight line be drawn - within a circle given in

magnitude cutting off a

to the angle in the

...

PROP . XCVIII . in jier , P. 2 IF a straight line be drawn - within a circle given in

magnitude cutting off a

**segment**containing a given angle ; if the angle adjacentto the angle in the

**segment**be bisećied by a straight line produced till it meet the...

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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.