## The First Six and the Eleventh and Twelfth Books of Euclid's Elements: With Notes and Illus., and an Appendix in Five Books |

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Side 67

because F is the centre of the

centre of the circle DBF, FB is equal to F.E. But FB was shown to be equal to FC:

therefore FE is equal (I. ax. 1.) to FC, which (I. ax. 9.) is impossible. Therefore F ...

because F is the centre of the

**circle ABC**, FB is equal to FC; and, because F is thecentre of the circle DBF, FB is equal to F.E. But FB was shown to be equal to FC:

therefore FE is equal (I. ax. 1.) to FC, which (I. ax. 9.) is impossible. Therefore F ...

Side 70

ONE circle cannot cut another in more than two points. If it be possible, let the

circle FAB cut the circle DEF in more than two points, viz., in B, G, F : take (III. 1.)

the centre H, of the

circle ...

ONE circle cannot cut another in more than two points. If it be possible, let the

circle FAB cut the circle DEF in more than two points, viz., in B, G, F : take (III. 1.)

the centre H, of the

**circle ABC**, and join HB, HG, H.F. Then, because within thecircle ...

Side 89

PROP. II. PROB. IN a given

triangle. Let

**ABC**, a chord is placed equal to the given straight line D; which was to be done.PROP. II. PROB. IN a given

**circle**to inscribe a triangle equiangular to a giventriangle. Let

**ABC**be a given**circle**, and DEF a given triangle; it is required to ... Side 268

Their boundaries are also equal, the boundary of each being equal to the

circumference of the circle.* PROP. XLII. PROB. To divide a given

any proposed number of equal parts, by means of concentric circles. Divide the

radius ...

Their boundaries are also equal, the boundary of each being equal to the

circumference of the circle.* PROP. XLII. PROB. To divide a given

**circle ABC**intoany proposed number of equal parts, by means of concentric circles. Divide the

radius ...

Side 269

Hence it is plain, that the area of any annulus, or ring, between the

circumferences of two concentric circles, such as that between the

circumferences of ABC and GI, is to the

squares of the radii DM, DL to the ...

Hence it is plain, that the area of any annulus, or ring, between the

circumferences of two concentric circles, such as that between the

circumferences of ABC and GI, is to the

**circle ABC**, as the difference of thesquares of the radii DM, DL to the ...

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The First Six and the Eleventh and Twelfth Books of Euclid's Elements: With ... Euclid Ingen forhåndsvisning tilgjengelig - 2016 |

### Vanlige uttrykk og setninger

ABCD altitude angle ABC angle BAC angle equal BC is equal bisected centre chord circle ABC circumference cone const contained cylinder describe a circle diagonal diameter divided draw equal angles equal to AC equiangular equilateral Euclid exterior angle fore fourth given circle given point given ratio given straight line greater half Hence hypotenuse inscribed join less Let ABC magnitudes manner multiple opposite parallel parallelepiped parallelogram perpendicular polygon polyhedron prism PROB produced PROP proportional proposition pyramid radius rectangle rectilineal figure right angles Schol segments semicircle sides similar similar triangles solid angles square of AC straight lines drawn tangent THEOR third triangle ABC triplicate ratio vertex vertical angle wherefore

### Populære avsnitt

Side 94 - If two triangles have two angles of the one equal to two angles of the other, each to each, and one side equal to one side, viz.

Side 53 - If a straight line be divided into any two parts, the squares of the whole line and of one of the parts are equal to twice the rectangle contained by the whole and that part, together with the square of the other part. Let the straight line AB be divided into any two parts at the point C : the squares of AB, BC shall be equal to twice the rectangle AB, BC, together with the square of AC.

Side 143 - 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 4 - A rhombus is that which has all its sides equal, but its angles are not right angles.

Side 57 - To divide a given straight line into two parts, so that the rectangle contained by the whole and one of the parts may be equal to the square on the other part.

Side 138 - IF a straight line be drawn parallel to one of the sides of a triangle, it shall cut the other sides, or those produced, proportionally; and if the sides, or the sides produced, be cut proportionally, the straight line which joins the points of section shall be parallel to the remaining side of the triangle...

Side 43 - In any right-angled triangle, the square which is described upon the side subtending the right angle, is equal to the squares described upon the sides which contain the right angle.

Side 32 - All the interior angles of any rectilineal figure, together with four right angles, are equal to twice as many right angles as the figure has sides.

Side 40 - 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.

Side 36 - PARALLELOGRAMS upon the same base, and between the same parallels, are equal to one another...