## The Elements of Euclid |

### Inni boken

Resultat 1-5 av 16

Side 33

Produce AD both ways to the points G , H , and through B

1 . ) to CA , and through F

GBCA , DEFH is a parallelogram ; and they are equal ( 36 . 1 . ) to one another ...

Produce AD both ways to the points G , H , and through B

**draw**BG parallel ( 31 .1 . ) to CA , and through F

**draw**FH parallel to ED : then each of the figures G HGBCA , DEFH is a parallelogram ; and they are equal ( 36 . 1 . ) to one another ...

Side 43

the square CEFB , join BE , and through D

BF ; and through H

parallel to CL or BM ; and because the complement CH is equal ( 43 . 1 . ) to the ...

the square CEFB , join BE , and through D

**draw**( 31 . 1 . ) DHG parallel to CE orBF ; and through H

**draw**KLM parallel to CB or EF ; and also through A**draw**AKparallel to CL or BM ; and because the complement CH is equal ( 43 . 1 . ) to the ...

Side 65

the centre E of the circle , and join AE ; and from the centre E , at the distance EA ,

describe the circle AFG ; from the point D

, and join EBF , AB . AB touches the circle BCD . Because E is the centre of the ...

the centre E of the circle , and join AE ; and from the centre E , at the distance EA ,

describe the circle AFG ; from the point D

**draw**( 11 . 1 . ) DF at right angles to EA, and join EBF , AB . AB touches the circle BCD . Because E is the centre of the ...

Side 89

Let ABCDE be the given equilateral and equiangular pentagon : it is required to

inscribe a circle in the pentagon ABCDE . Bisect ( 9 . 1 . ) the angles BCD , CDE

by the straight lines CF , DF , and from the point F , in which they meet ,

...

Let ABCDE be the given equilateral and equiangular pentagon : it is required to

inscribe a circle in the pentagon ABCDE . Bisect ( 9 . 1 . ) the angles BCD , CDE

by the straight lines CF , DF , and from the point F , in which they meet ,

**draw**the...

Side 163

To

A be the given point above the plane BH ; it is required to

straight line perpendicular to the plane BH . In the plane

To

**draw**a straight line perpendicular to a plane , from a given point above it . LetA be the given point above the plane BH ; it is required to

**draw**from the point A astraight line perpendicular to the plane BH . In the plane

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### Vanlige uttrykk og setninger

added altitude angle ABC angle BAC base BC is given 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 magnitude given in position given in species given magnitude given ratio given straight line gles greater Greek half join less likewise magnitude manner meet multiple Note opposite parallel parallelogram pass perpendicular plane prism produced PROP proportionals proposition pyramid Q. E. D. PROP reason rectangle rectangle contained remaining right angles segment shown sides similar sine solid sphere square square of BC taken THEOR third triangle ABC wherefore whole

### Populære avsnitt

Side 36 - 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 145 - Wherefore, in equal circles &c. QED PROPOSITION B. THEOREM If the vertical angle of a triangle be bisected by a straight line which likewise cuts the base, the rectangle contained by the sides of the triangle is equal to the rectangle contained by the segments of the base, together with the square on the straight line which bisects the angle.

Side 65 - The angle in a semicircle is a right angle; the angle in a segment greater than a semicircle is less than a right angle; and the angle in a segment less than a semicircle is greater than a right angle.

Side 248 - 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 11 - If one side of a triangle be produced, the exterior angle is greater than either of the interior opposite angles.

Side 121 - 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 21 - 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 14 - To construct a triangle of which the sides shall be equal to three given straight lines ; but any two whatever of these lines must be greater than the third (20.

Side 80 - 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. either the sides adjacent to the equal...

Side 133 - ... rectilineal figures are to one another in the duplicate ratio of their homologous sides.