Euclid's Elements of geometry, books i. ii. iii. iv1862 |
Inni boken
Resultat 1-5 av 30
Side 38
... passing through A can be parallel to BC , except AD ; 6. Therefore AD is parallel to BC . Conclusion . Therefore , equal triangles , & c . Q. E. D. PROPOSITION 40. - THEOREM . Equal triangles upon the same side of equal bases , that are ...
... passing through A can be parallel to BC , except AD ; 6. Therefore AD is parallel to BC . Conclusion . Therefore , equal triangles , & c . Q. E. D. PROPOSITION 40. - THEOREM . Equal triangles upon the same side of equal bases , that are ...
Side 39
... passing through A , can be parallel to BF , except AD ; 6. Therefore AD is parallel to BF . Conclusion . Therefore , equal triangles , & c . Q. E. D. PROPOSITION 41. - THEOREM . If a parallelogram and a triangle be upon the same base ...
... passing through A , can be parallel to BF , except AD ; 6. Therefore AD is parallel to BF . Conclusion . Therefore , equal triangles , & c . Q. E. D. PROPOSITION 41. - THEOREM . If a parallelogram and a triangle be upon the same base ...
Side 40
... passes ; and BK , KD the other paral- lelograms , which make up the whole figure ABCD , and are therefore called the complements . Sequence . The complement BK shall be equal to the com- plement KD . Demonstration . - 1 . Because ABCD ...
... passes ; and BK , KD the other paral- lelograms , which make up the whole figure ABCD , and are therefore called the complements . Sequence . The complement BK shall be equal to the com- plement KD . Demonstration . - 1 . Because ABCD ...
Side 48
... passing through two given points , and having its centre in a given straight line . 12. If two adjacent angles be bisected , the bisecting lines contain a right angle . Prop . 16 to 26 . 13. Prove that only one perpendicular can be ...
... passing through two given points , and having its centre in a given straight line . 12. If two adjacent angles be bisected , the bisecting lines contain a right angle . Prop . 16 to 26 . 13. Prove that only one perpendicular can be ...
Side 50
... pass through the same point , and trisect each other in that point . 54. Given the straight line drawn from the vertex of an equilateral triangle to one of the points of trisection of the base , to construct the triangle . BOOK II ...
... pass through the same point , and trisect each other in that point . 54. Given the straight line drawn from the vertex of an equilateral triangle to one of the points of trisection of the base , to construct the triangle . BOOK II ...
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.