Euclid's Elements of Geometry, Bøker 1-6Henry Martyn Taylor The University Press, 1893 - 504 sider |
Inni boken
Resultat 1-5 av 83
Side 16
... triangle . Let AB be the given finite straight line : it is required to construct an equilateral triangle on AB ... ABC is a triangle constructed as required . A B E PROOF . Because A is the centre of the circle BCD , AC is equal to AB ...
... triangle . Let AB be the given finite straight line : it is required to construct an equilateral triangle on AB ... ABC is a triangle constructed as required . A B E PROOF . Because A is the centre of the circle BCD , AC is equal to AB ...
Side 22
... ABC , DEF be two triangles , in which AB is equal to DE , and AC to DF , and the angle BAC is equal to the angle EDF ... triangle ABC so that A coincides with D , and AB coincides in direction with DE , and AC with DF . ( Test of ...
... ABC , DEF be two triangles , in which AB is equal to DE , and AC to DF , and the angle BAC is equal to the angle EDF ... triangle ABC so that A coincides with D , and AB coincides in direction with DE , and AC with DF . ( Test of ...
Side 23
... triangle ABC is shifted , so that A is on D and AB is on DE , there would be no justification for the conclusion that B must coincide with E , because AB is equal to DE , if it were possible for two straight lines to have a common part ...
... triangle ABC is shifted , so that A is on D and AB is on DE , there would be no justification for the conclusion that B must coincide with E , because AB is equal to DE , if it were possible for two straight lines to have a common part ...
Side 26
... ABC be a triangle , in which the angle ABC is equal to the angle ACB : it is required to prove that AC is equal to AB . CONSTRUCTION . Let the triangle ABC be turned over and shifted unchanged in shape and size to the position . abc , A ...
... ABC be a triangle , in which the angle ABC is equal to the angle ACB : it is required to prove that AC is equal to AB . CONSTRUCTION . Let the triangle ABC be turned over and shifted unchanged in shape and size to the position . abc , A ...
Side 27
... ABC and ACB at the base of an isosceles triangle be bisected by the straight lines BD and CD , DBC will be an isosceles triangle . 2. BAC is a triangle having the angle B double of the angle A. If BD bisect the angle B and meet AC at D ...
... ABC and ACB at the base of an isosceles triangle be bisected by the straight lines BD and CD , DBC will be an isosceles triangle . 2. BAC is a triangle having the angle B double of the angle A. If BD bisect the angle B and meet AC at D ...
Andre utgaver - Vis alle
Vanlige uttrykk og setninger
ABCD AC is equal ADDITIONAL PROPOSITION angle ACB angle BAC angles ABC anharmonic arc ABC bisected centre of similitude chord circle ABC coincide Constr Coroll cut the circle describe a circle diagonal diameter draw equal angles equal circles equal to CD equiangular equimultiples Euclid EXERCISES exterior angle given circle given point given straight line given triangle greater harmonic range hypotenuse inscribed intersect Let ABC meet middle points opposite sides pair parallel parallelogram pencil pentagon perpendicular polygon PROOF Prop PROPOSITION 14 Ptolemy's Theorem quadrilateral radical axis radius rectangle contained required to prove respectively rhombus right angles shew sides BC Similarly square on AC straight line &c straight line drawn straight line joining subtend tangent theorem triangle ABC triangle DEF triangles are equal twice the rectangle vertices Wherefore
Populære avsnitt
Side 59 - Any two sides of a triangle are together greater than the third side.
Side 7 - An angle less than a right angle is called an acute angle; an angle greater than a right angle and less than two right angles is called an obtuse angle.
Side 68 - 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 144 - 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 376 - To find a mean proportional between two given straight lines. Let AB, BC be the two given straight lines ; it is required to find a mean proportional between them. Place AB, BC in a straight line, and upon AC describe the semicircle ADC, and from the point B draw (9.
Side 135 - If there be two straight lines, one of which is divided into any number of parts, the rectangle contained by the two straight lines is equal to the rectangles contained by the undivided line, and the several parts of the divided line.
Side 76 - ... the same side together equal to two right angles ; the two straight lines shall be parallel to one another.
Side 305 - To inscribe, an equilateral and equiangular pentagon in a given circle. Let ABCDE be the given circle. It is required to inscribe an equilateral...
Side 424 - PROPOSITION 5. The locus of a point, the ratio of whose distances from two given points is constant, is a circle*.
Side 248 - If two straight lines within a circle cut one another, the rectangle contained by the segments of one of them is equal to the rectangle contained by the segments of the other. Let the two straight lines AC, BD, within the circle ABCD, cut one another in the point E : the rectangle contained by AE, EC is equal to the rectangle contained by BE, ED.