## Euclid's Elements of geometry, books i. ii. iii. iv |

### Inni boken

Resultat 1-5 av 5

Side 17

But the angles CEA , AED , have been

angles . 4 . Therefore the angles CEĂ , AED , are equal to the angles AED , DEB .

( ax . 1 . ) 5 . Take away the common angle AED , 6 . The remaining CEA is equal

...

But the angles CEA , AED , have been

**shown**to be together equal to two rightangles . 4 . Therefore the angles CEĂ , AED , are equal to the angles AED , DEB .

( ax . 1 . ) 5 . Take away the common angle AED , 6 . The remaining CEA is equal

...

Side 116

In the same manner it may be

T C K to BD ; 8 . Therefore the figures GK , GC , CF , FB , BK , are parallelograms ;

9 . Therefore GF is equal to HK , and GH to FK . ( I . 34 . ) 10 . And because AC is

...

In the same manner it may be

**shown**that each of the lines GF , HK , is parallelT C K to BD ; 8 . Therefore the figures GK , GC , CF , FB , BK , are parallelograms ;

9 . Therefore GF is equal to HK , and GH to FK . ( I . 34 . ) 10 . And because AC is

...

Side 122

And the base BK was

angle BFK is equal to the angle CFK ( I . 8 ) ; and the angle BKF to the angle CKF

; ( I . 4 . ) 13 . Therefore the angle BFC is double of the angle CFK , and the angle

...

And the base BK was

**shown**to be equal to the base CK : 12 . Therefore theangle BFK is equal to the angle CFK ( I . 8 ) ; and the angle BKF to the angle CKF

; ( I . 4 . ) 13 . Therefore the angle BFC is double of the angle CFK , and the angle

...

Side 125

In the same manner it may be

FC or FD ; , 6 . Therefore the five straight lines FA , FB , FC , FD , FE , are equal to

one another , and the circle described from the centre F , at the distance of any ...

In the same manner it may be

**shown**that FB , FA , FE , are each of them equal toFC or FD ; , 6 . Therefore the five straight lines FA , FB , FC , FD , FE , are equal to

one another , and the circle described from the centre F , at the distance of any ...

Side 126

In the same manner it may be

ABCDEF are each of them equal to the angle AFE or FED ; 24 . Therefore the

hexagon is equiangular ; and it was

Therefore , a ...

In the same manner it may be

**shown**that the other angles of the hexagonABCDEF are each of them equal to the angle AFE or FED ; 24 . Therefore the

hexagon is equiangular ; and it was

**shown**to be equilateral . Conclusion .Therefore , a ...

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### Vanlige uttrykk og setninger

ABCD angle ABC angle BAC angle BCD angle equal assumed base base BC BC is equal bisected BOOK centre circle ABC circumference coincide common Conclusion Conclusion.—Therefore const Construction Construction.-1 Demonstration Demonstration.-1 describe diameter distance divided double draw drawn equilateral exterior angle extremities fall figure four given circle given point given rectilineal given straight line greater half Hypothesis.—Let inscribed join less manner meet opposite angle parallel parallelogram pass pentagon perpendicular produced Prop proved Q. E. D. PROPOSITION reason rectangle contained rectilineal figure References—Prop regular right angles segment semicircle shown sides Sought.-It is required square on AC Take third touches the circle triangle ABC twice the rectangle whole

### Populære avsnitt

Side 25 - 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 2 - A circle is a plane figure contained by one line, which is called the circumference, and is such that all straight lines drawn from a certain point within the figure to the circumference, are equal to one another.

Side 99 - 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 4 - 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 66 - ... the sides containing the obtuse angle, by twice the rectangle contained by the side upon which, when produced, the perpendicular falls, and the straight line intercepted without the triangle between the perpendicular and the obtuse...

Side 65 - 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 of the other part.

Side 32 - F, which is the common vertex of the triangles ; that is, together with four right angles. Therefore all the angles of the figure, together with four right angles, are equal to twice as many right angles as the figure has sides.

Side 58 - If a straight line be bisected, and produced to any point ; the rectangle contained by the whole line thus produced, and the part of it produced...

Side 88 - The straight line drawn at right angles to the diameter of a circle, from the extremity of it, falls without the circle...

Side 33 - The straight lines which join the extremities of two equal and parallel straight lines towards the same parts, are also themselves equal and parallel.