## 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 186

Let ABC be a

AD , the

described about ...

Let ABC be a

**triangle**, and let the angle BAC be bisected by by the straight lineAD , the

**rectangle**BA , AC is**equal to the**...**of the triangle is equal to the****rectangle contained by the perpendicular and the dia**. meter of the**circle**described about ...

Side 439

that fpace be

must necessarily be

and its sides AB , BC are those which were to be

AC and draw BE

describe the

FB in HL pofiК. tion 8. and the

8 31.

that fpace be

**equal**to the given sum of the squares , the sides of the**rectangle**must necessarily be

**equal**to one ... a square ABCD**equal**to the given**rectangle**,and its sides AB , BC are those which were to be

**found**. for the**rectangle**AC is ...AC and draw BE

**perpendicular**to it , and complete the**rectangle**AEBF , anddescribe the

**circle**ABC aboạt the**triangle**... Dat . is given , and the straight lineFB in HL pofiК. tion 8. and the

**circumference**ABC is given in position , wherefore8 31.

Side 34

WHE

fines of two arches , is

center of the

likewise

WHE

**rectangle contained**by half of the radius , and the excess of the versedfines of two arches , is

**equal to the rectangle contained**... AB ; through E thecenter of the

**circle**, let there be drawn a**diameter**DEF , and AE joined , and CDlikewise

**perpendicular**to it in G ; and ... KCG will be**equal**; but KCG is in a**semicircle**, and therefore a right angle ; therefore BKC is a right angle ; and**in the****triangles**DFB ...### 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.