## 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 104

... same

or AED . therefore the pentagon ABCDE is equiangular ; and it has been shewn

that it is equilateral . Wherefore in the given circle an equilateral and equiangular

...

... same

**reason**, each of the angles ABC , BCD , CDE is equal to the angle BAEor AED . therefore the pentagon ABCDE is equiangular ; and it has been shewn

that it is equilateral . Wherefore in the given circle an equilateral and equiangular

...

Side 169

So the equal angles EBC , LGH are proportionals ; therefore the triangle EBC is

equiangular to the triangle LGH , and similar to it c . for the fame

А gle ECD like , M wife is similar to the triangle E LHK . therefore the similar poly

...

So the equal angles EBC , LGH are proportionals ; therefore the triangle EBC is

equiangular to the triangle LGH , and similar to it c . for the fame

**reason**the trian ,А gle ECD like , M wife is similar to the triangle E LHK . therefore the similar poly

...

Side 200

Def . 11. EF , it is perpendicular a to the straight C H line BK which is in that plane

. thereF fore ABK is a right angle . for the same A

wherefore the two angles ABK , BAK of the triangle ABK are equal to two right b .

Def . 11. EF , it is perpendicular a to the straight C H line BK which is in that plane

. thereF fore ABK is a right angle . for the same A

**reason**, BAK is a right angle ;wherefore the two angles ABK , BAK of the triangle ABK are equal to two right b .

Side 250

Book XII . equal to the bafe KD , and the triangle AEH equal d and similar to the

triangle HKD . for the same

triangle HLD . and because the two straight lines EH , HG which meet one

another ...

Book XII . equal to the bafe KD , and the triangle AEH equal d and similar to the

triangle HKD . for the same

**reason**, the triangle AGH is equal and similar to thetriangle HLD . and because the two straight lines EH , HG which meet one

another ...

Side 308

Clavius likewise gives , both the ways , but neither he nor Peletarius takes notice

of the

cases of this Proposition , of which only the first and simplest is demonstrated in ...

Clavius likewise gives , both the ways , but neither he nor Peletarius takes notice

of the

**reason**why one is preferable to the other . PROP . VI . B. V. There are twocases of this Proposition , of which only the first and simplest is demonstrated in ...

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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.