## The first three books of Euclid's Elements of geometry, with theorems and problems, by T. Tate |

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Resultat 1-5 av 5

Side 24

lines AB , CD , make the alternate angles A EF , EFD equal to one another ; AB is

either towards B , D , or towards A , C ; let them be produced and meet towards B

...

lines AB , CD , make the alternate angles A EF , EFD equal to one another ; AB is

**parallel**to CD . For , if it be not**parallel**, AB and cd being produced , shall meeteither towards B , D , or towards A , C ; let them be produced and meet towards B

...

Side 26

H н F Let the straight line Gak cut AB , EF , CD ; and because GHK cuts the

GHF . Again , because the straight line gak cuts the

CD ...

H н F Let the straight line Gak cut AB , EF , CD ; and because GHK cuts the

**parallel**straight lines A B , EF , the angle Agu is equal ( I. 29. ) A. B to the angleGHF . Again , because the straight line gak cuts the

**parallel**E. straight lines EF ,CD ...

Side 28

The straight lines which join the extremities of two equal and

lines , towards the same parts , are also themselves equal and

CD be equal and

the ...

The straight lines which join the extremities of two equal and

**parallel**straightlines , towards the same parts , are also themselves equal and

**parallel**. Let A B ,CD be equal and

**parallel**straight lines , and joined towards the same parts bythe ...

Side 32

A E

triangle EBC , because it is upon the same base BC , and between the A D same

parallels BC , AE : But the triangle ABC is equal to the triangle BDC ( Hyp . ) ;

therefore ...

A E

**parallel**to BC , and join EC : The triangle ABC is equal ( 1. 37. ) to thetriangle EBC , because it is upon the same base BC , and between the A D same

parallels BC , AE : But the triangle ABC is equal to the triangle BDC ( Hyp . ) ;

therefore ...

Side 92

FG

Fg is

GFA ; but ( by construction ) the angle EAF is equal to the angle gaF ; therefore

the ...

FG

**parallel**to EA , and**parallel**to KB ; then ghy is the triangle required . For sinceFg is

**parallel**to EA , therefore ( E. 1. 29. ) the angle EAF is equal to the angleGFA ; but ( by construction ) the angle EAF is equal to the angle gaF ; therefore

the ...

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### Andre utgaver - Vis alle

The first three books of Euclid's Elements of geometry, with theorems and ... Euclides Uten tilgangsbegrensning - 1851 |

The First Three Books of Euclid's Elements of Geometry from the Text of Dr ... Euclid,Thomas Tate Ingen forhåndsvisning tilgjengelig - 2014 |

The First Three Books of Euclid's Elements of Geometry from the Text of Dr ... Euclid,Thomas Tate Ingen forhåndsvisning tilgjengelig - 2014 |

### Vanlige uttrykk og setninger

ABCD alternate angle ABC angle ACB angle BAC angle equal base BC BC is equal bisect centre circle ABC circumference coincide common construct demonstrated describe diameter divided double draw equal angles equal to FB equilateral exterior angle extremity figure fore four given point given straight line gnomon greater impossible interior isosceles triangle join less Let ABC likewise line be drawn meet opposite angles opposite sides parallel parallelogram pass perpendicular PROB produced PROP Q. E. D. PROP rectangle contained remaining angle right angles segment semicircle shown sides squares of AC straight line AC Take taken THEOR third touch touches the circle triangle ABC twice the rectangle vertex wherefore whole

### Populære avsnitt

Side 6 - If a straight line meets two straight lines, so as to make the two interior angles on the same side of it taken together less than two right angles...

Side 5 - Let it be granted that a straight line may be drawn from any one point to any other point.

Side 20 - If two triangles have two sides of the one equal to two sides of the...

Side 30 - Parallelograms upon equal bases, and between the same parallels, are equal to one another.

Side 17 - Any two angles of a triangle are together less than two right angles. Let ABC be any triangle ; any two of its angles together are less than two right angles.

Side 84 - IF from a point without a circle there be drawn two straight lines, one of which cuts the circle, and the other meets it; if the rectangle contained by the whole line which cuts the circle, and the part of it without the circle, be equal to the square of the line which meets it, the line which meets it shall touch the circle.

Side 82 - If from any point without a circle two straight lines be drawn, one of -which cuts the circle, and the other touches it; the rectangle contained by the whole line which cuts the circle, and the part of it without the circle, shall be equal to the square of the line which touches it.

Side 11 - UPON the same base, and on the same side of it, there cannot be two triangles that have their sides which are terminated in one extremity of the base equal to one another, and likewise those which are terminated in the other extremity.

Side 19 - To make a triangle of which the sides shall be equal to three given straight lines, but any two whatever of these must be greater than the third, (i.

Side 7 - From the greater of two given straight lines to cut off a part equal to the less. Let AB and C be the two given straight lines, whereof AB is the greater.