Euclid's Elements of geometry, books i. ii. iii. iv |
Inni boken
Resultat 1-5 av 52
Side 12
3. Join DE . 4. Upon DE , on the side remote from A , describe an equilateral triangle DEF . ( I. 1. ) 5. Join AF . DA B F Then the straight line AF shall bisect the angle BAC . Proof . - 1 . Because AD is equal to AE ( const . ) ...
3. Join DE . 4. Upon DE , on the side remote from A , describe an equilateral triangle DEF . ( I. 1. ) 5. Join AF . DA B F Then the straight line AF shall bisect the angle BAC . Proof . - 1 . Because AD is equal to AE ( const . ) ...
Side 13
And the base DF is equal to the base EF ; ( const . ) 4. Therefore the angle DAF is equal to the angle EAF . ( I.8 . ) Conclusion . Therefore the given rectilineal angle BÀC is bisected by the straight line AF . Q. E. F. PROPOSITION 10.
And the base DF is equal to the base EF ; ( const . ) 4. Therefore the angle DAF is equal to the angle EAF . ( I.8 . ) Conclusion . Therefore the given rectilineal angle BÀC is bisected by the straight line AF . Q. E. F. PROPOSITION 10.
Side 14
Because DC is equal to CE ( const . ) , and FC common to the two triangles DCF , ECF ; 2. The two sides DC , CF , are equal to the two sides EC , CF , each to each ; 3. And the base DF is equal to the base EF ; ( const . ) 4.
Because DC is equal to CE ( const . ) , and FC common to the two triangles DCF , ECF ; 2. The two sides DC , CF , are equal to the two sides EC , CF , each to each ; 3. And the base DF is equal to the base EF ; ( const . ) 4.
Side 15
Because FH is equal to HG ( const . ) , and HC common to the two triangles FHC , GHC ; D GB 2. The two sides FH , HC , are equal to the two sides GH , HC , each to each ; 3. And the base CF is equal to the base CG ; ( def . 15. ) 4.
Because FH is equal to HG ( const . ) , and HC common to the two triangles FHC , GHC ; D GB 2. The two sides FH , HC , are equal to the two sides GH , HC , each to each ; 3. And the base CF is equal to the base CG ; ( def . 15. ) 4.
Side 18
1 . Bisect AC in E. ( I. 10. ) 2. Join BE , and produce it to F , making EF equal to BE ( I. 3 ) , and join FC . A A B R D Demonstration . - 1 . Because AE is equal to EC , and BE equal to EF , ( const . ) 2.
1 . Bisect AC in E. ( I. 10. ) 2. Join BE , and produce it to F , making EF equal to BE ( I. 3 ) , and join FC . A A B R D Demonstration . - 1 . Because AE is equal to EC , and BE equal to EF , ( const . ) 2.
Hva folk mener - Skriv en omtale
Vi har ikke funnet noen omtaler på noen av de vanlige stedene.
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 const Construction Construction.-1 Demonstration Demonstration.-1 describe diameter distance divided double draw drawn equal equilateral exterior angle extremities fall figure four given circle given point given straight line Given.-Let greater half Hypothesis Hypothesis.-Let inscribed join less manner meet opposite angle parallel parallelogram pass pentagon perpendicular produced proved Q. E. D. PROPOSITION reason rectangle AB BC rectangle contained References-Prop regular right angles segment semicircle Sequence shown sides Sought 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.
Bibliografisk informasjon
| Tittel | Euclid's Elements of geometry, books i. ii. iii. iv Scot. sch. book assoc. Progressive series |
| Forfatter | Euclides |
| distribuert | 1862 |
| Original fra | Oxford University |
| Digitalisert | 19. apr 2006 |
| Eksporter sitat | BiBTeX EndNote RefMan |