Euclid's Elements of geometry, books i. ii. iii. iv1862 |
Inni boken
Resultat 1-5 av 34
Side 8
... take any point F. 2. From AE , the greater , cut off AG , equal to AF , the less . ( I. 3. ) 3. Join FC , GB . D F B A E Demonstration . - 1 . Because AF is equal to AG . ( constr . ) And AB is equal to AC . ( hypoth . ) Therefore the ...
... take any point F. 2. From AE , the greater , cut off AG , equal to AF , the less . ( I. 3. ) 3. Join FC , GB . D F B A E Demonstration . - 1 . Because AF is equal to AG . ( constr . ) And AB is equal to AC . ( hypoth . ) Therefore the ...
Side 12
... Take any point D in AB . 2. From AC cut off AE equal to AD . ( I. 3. ) 3. Join DE . 4. Upon DE , on the side remote from A , describe an equilateral triangle DEF . ( I. 1. ) 5. Join AF . B D Then the straight line AF shall bisect the ...
... Take any point D in AB . 2. From AC cut off AE equal to AD . ( I. 3. ) 3. Join DE . 4. Upon DE , on the side remote from A , describe an equilateral triangle DEF . ( I. 1. ) 5. Join AF . B D Then the straight line AF shall bisect the ...
Side 13
... Take any point D in AC . 2. Make CE equal to CD . ( I. 3. ) 3. Upon DE describe the equilateral triangle DFE . ( I. Į . ) 4. Join FC . Then FC shall be at right angles to AB . Proof . - 1 . Because DC is equal to BOOK I. PROP . 10 . 13.
... Take any point D in AC . 2. Make CE equal to CD . ( I. 3. ) 3. Upon DE describe the equilateral triangle DFE . ( I. Į . ) 4. Join FC . Then FC shall be at right angles to AB . Proof . - 1 . Because DC is equal to BOOK I. PROP . 10 . 13.
Side 15
... Take any point D upon the other side of AB . 2. From the centre C , at the dis- tance CD , describe the circle EGF , meeting AB in F and G. ( post . 3. ) 3. Bisect FG in H. ( I. 10. ) 4. Join CF , CH , CG . H E A F G B D Then CH shall ...
... Take any point D upon the other side of AB . 2. From the centre C , at the dis- tance CD , describe the circle EGF , meeting AB in F and G. ( post . 3. ) 3. Bisect FG in H. ( I. 10. ) 4. Join CF , CH , CG . H E A F G B D Then CH shall ...
Side 17
... Take away the common angle ABC , 6. The remaining angle ABE is equal to the remaining angle ABD ( ax . 3 ) , the less to the greater ; which is impossible . 7. Therefore BE is not in the same straight line with BC . 8. And , in like ...
... Take away the common angle ABC , 6. The remaining angle ABE is equal to the remaining angle ABD ( ax . 3 ) , the less to the greater ; which is impossible . 7. Therefore BE is not in the same straight line with BC . 8. And , in like ...
Vanlige uttrykk og setninger
AB is equal AC and CB adjacent angles angle ABC angle AGH angle BAC angle BCD angle EAB angle EDF angle equal angles CBA base BC BC is equal bisected circle ABC circumference Conclusion Conclusion.-Therefore const Construction.-1 Demonstration.-1 describe the circle diameter double equal angles equal to CD equiangular exterior angle given circle given point given rectilineal angle given straight line Given.-Let ABCD gnomon greater Hypothesis inscribed interior and opposite isosceles triangle less opposite angle parallel to CD parallelogram perpendicular point F produced Q. E. D. PROPOSITION rectangle AB BC rectangle AE rectangle contained rectilineal figure References-Prop remaining angle required to describe right angles segment semicircle Sequence side BC square on AC straight line AC straight line drawn touches the circle triangle ABC triangle DEF twice the rectangle
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.