## The Elements of Euclid: The Errors, by which Theon, Or Others, Have Long Ago Vitiated These Books are Corrected, and Some of Euclid's Demonstrations are Restored. Also, the Book of Euclid's Data, in Like Manner Corrected. viz. the first six books, together with the eleventh and twelfth |

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Resultat 1-5 av 5

Side 24

base EG ; and because DG is equal to DF , the

**BAC**is equal to the**angle**A D EDG ; therefore the base BC is equal ( 4. 1. ) to thebase EG ; and because DG is equal to DF , the

**angle**DFG is equal ( 5. 1. ) to the**angle**DGF ; but the**angle**DGF is greater than the**angle**EGF ; E G therefore the ... Side 73

Let ABCD be a circle , of which the diameter is BC , and centre E ; and draw CA ,

dividing the circle into the segments ABC , ADC , and join BA , AD , DC ; the

Let ABCD be a circle , of which the diameter is BC , and centre E ; and draw CA ,

dividing the circle into the segments ABC , ADC , and join BA , AD , DC ; the

**angle**in the semicircle**BAC**is a right**angle**; and the**angle**in the segment ABC ... Side 169

Wherefore , if a solid angle , & c . Q. E. D. B PROP . XXI . THEOR . Every solid

angle is contained by plane angles which together are less than four right angles

. First , let the solid angle at A be contained by three plane

Wherefore , if a solid angle , & c . Q. E. D. B PROP . XXI . THEOR . Every solid

angle is contained by plane angles which together are less than four right angles

. First , let the solid angle at A be contained by three plane

**angles BAC**, CAD ... Side 317

to the

angles at B , C are given . And because the sides AB , BC , CA are given , their ...

to the

**angle BAC**; therefore , because the angle EDF which is equal to the**angle****BAC**, has been found , the**angle BAC**is given ( 1. def . ) ; in like manner theangles at B , C are given . And because the sides AB , BC , CA are given , their ...

Side 347

to FL , must not be greater than the ratio of MN to BC : let it be so ; and , by the

77th dat . find a triangle OPQ which has one of its angles POQ equal to the given

to FL , must not be greater than the ratio of MN to BC : let it be so ; and , by the

77th dat . find a triangle OPQ which has one of its angles POQ equal to the given

**angle BAC**, and the ratio of the perpendicular OR , drawn from that angle to the ...### Hva folk mener - Skriv en omtale

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The Elements of Euclid, Viz; The First Six Books: Together With the Eleventh ... Euclid Euclid Ingen forhåndsvisning tilgjengelig - 2018 |

The Elements of Euclid: Viz. The First Six Books, Together With the Eleventh ... Euclid Euclid Ingen forhåndsvisning tilgjengelig - 2017 |

### Vanlige uttrykk og setninger

added altitude angle ABC angle BAC base centre circle circle ABCD circumference common cone contained cylinder definition demonstrated described diameter divided double draw drawn equal equal angles equiangular equimultiples Euclid excess fore four fourth given angle given in position given in species given magnitude given ratio given straight line gles greater Greek half join less likewise manner meet multiple Note opposite parallel parallelogram pass perpendicular plane prism produced PROP proportionals proposition pyramid Q. E. D. PROP radius reason rectangle rectangle contained rectilineal figure remaining right angles segment shown sides similar sine solid solid angle sphere square square of AC taken THEOR third triangle ABC wherefore whole

### Populære avsnitt

Side 45 - Ir a straight line be divided into any two parts, four times the rectangle contained by the whole line, and one of the parts, together with the square of the other part, is equal to the square of the straight line which is made up of the whole and that part.

Side 41 - 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 54 - Ir any two points be taken in the circumference of a circle, the straight line which joins them shall fall within the circle. Let ABC be a circle, and A, B any two points in the circumference ; the straight line drawn from A to B shall fall within the circle.

Side 18 - ABD, the less to the greater, which is impossible ; therefore BE is not in the same straight line with BC.

Side 10 - From a given point to draw a straight line equal to a given straight line. Let A be the given point, and BC the given straight line: it is required to draw from the point A a straight line equal to BC.

Side 8 - Let it be granted that a straight line may be drawn from any one point to any other point.

Side 256 - Again ; the mathematical postulate, that " things which are equal to the same are equal to one another," is similar to the form of the syllogism in logic, which unites things agreeing in the middle term.

Side 129 - 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 23 - At a given point in a given straight line, to make a rectilineal angle equal to a given rectilineal angle. Let AB be the given straight line, and A...

Side 20 - ANY two angles of a triangle are together less than two right angles.