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

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Resultat 1-5 av 5

Side 57

If a straight line be divided into any two parts , the squares on the whole line , and

on one of the parts , are equal to

that part , together with the squirre on the other part . H ( References Prop .

If a straight line be divided into any two parts , the squares on the whole line , and

on one of the parts , are equal to

**twice the rectangle**contained by the whole andthat part , together with the squirre on the other part . H ( References Prop .

Side 58

CK and HF , is equal to

AC . 10 . But the gnomon AKF , together with the squares CK and HF , make up

the whole figure ADEB and CK , which are the squares on AB and BC . 11 .

CK and HF , is equal to

**twice the rectangle**AB . BC , together with the square onAC . 10 . But the gnomon AKF , together with the squares CK and HF , make up

the whole figure ADEB and CK , which are the squares on AB and BC . 11 .

Side 64

Therefore the remaining rectangle CF . ... But the figure FK is the rectangle

contained by CF and FA , for FA is equal to FG ; ( def . 30 . ) 10 . ... The square on

AB shall be greater than the squares on AC and CB , by

CD .

Therefore the remaining rectangle CF . ... But the figure FK is the rectangle

contained by CF and FA , for FA is equal to FG ; ( def . 30 . ) 10 . ... The square on

AB shall be greater than the squares on AC and CB , by

**twice the rectangle**BC .CD .

Side 65

Therefore the squares on BD and DA are equal to the squares on BC , CD , DA ,

and

on BD and DA , because the angle at D is a right angle ; ( I . 47 . ) 6 . And the ...

Therefore the squares on BD and DA are equal to the squares on BC , CD , DA ,

and

**twice the rectangle**BC . CD . 5 . But the square on BA is equal to the squareson BD and DA , because the angle at D is a right angle ; ( I . 47 . ) 6 . And the ...

Side 66

square on AC alone is less than the squares on CB and BA , by

Demonstration . - l . Because the angle at D is a right angle ( const . ) , the angle

ACB is ...

square on AC alone is less than the squares on CB and BA , by

**twice the****rectangle**CB . BD . Case II . - Secondly , let AD fall without the triangle ABC .Demonstration . - l . Because the angle at D is a right angle ( const . ) , the angle

ACB is ...

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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.