## 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

Side 24

Book

the

Book

**1**. Re 4. I. For if it be not greater , it must either be**equal**to it , or less . butthe

**angle**BAC is not**equal**to the**angle**... F**two**triangles have**two angles**of**one****equal to two angles of the other**; each to each , and**one**side**equal**to**one**side ... Side 31

I. gles , each to each , to which the

the

and makes , the alternate

c ...

I. gles , each to each , to which the

**equal**fides are opposite . there- Book I. forethe

**angle**ACB is**equal**to the**angle**... BC meets the '**two**straight lines AC , BDand makes , the alternate

**angles**ACB , CBD**equal**to**one another**, AC is parallelc ...

Side 97

14

be divided into

the

14

**1**quadrilateral figure AMBK are**equal**to four right**angles**, for it Book IV . canbe divided into

**two**triangles ; and that**two**of them KAM , KBM are right**angles**,the

**other two**AKB , AMB are**equal to two**right**angles**. but the L**angles**DEG ... Side 236

...

MDE there are

and

...

**angle**HAB is**equal**to the**angle**MDE . therefore in the**two**triangles HAB ,MDE there are

**two angles**in**one equal to two angles**in the**other**, each to each ,and

**one**side**equal**to**one**side , oppofite to**one**of the**equal angles**in each , viz . Side 242

YS and DG do meet , and cut

, SG . because DX is parallel to OE , the alC. 29. I. ternate

YS and DG do meet , and cut

**one another**into**two equal**parts . Join DY , YE , BS, SG . because DX is parallel to OE , the alC. 29. I. ternate

**angles**DXY , YOE are**equal**to**one another , and**becaufe DX is**equal**to OE , and XY to YO , D K F ...### 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.