Euclid's Elements of Geometry, Bøker 1-6Henry Martyn Taylor The University Press, 1893 - 504 sider |
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Resultat 1-5 av 87
Side 20
... equal to CD ; ( Construction . ) therefore AF is equal to CD . Wherefore , from AB the greater of two given straight lines a part AF has been cut off equal to CD the less . The demand made in Postulate 6 , that " a 20 BOOK I.
... equal to CD ; ( Construction . ) therefore AF is equal to CD . Wherefore , from AB the greater of two given straight lines a part AF has been cut off equal to CD the less . The demand made in Postulate 6 , that " a 20 BOOK I.
Side 26
... and is equal to it . But ac is equal to AČ ; therefore AC is equal to AB . Wherefore if two angles & c . COROLLARY . An equiangular triangle is also equilateral . When in two propositions the hypothesis of each is the 26 BOOK I.
... and is equal to it . But ac is equal to AČ ; therefore AC is equal to AB . Wherefore if two angles & c . COROLLARY . An equiangular triangle is also equilateral . When in two propositions the hypothesis of each is the 26 BOOK I.
Side 29
... ) ( 3 ) Next let the vertex D of one triangle lie on one of the sides BC of the other : then BC , BD are unequal . Wherefore , if two points on the same side & c . PROPOSITION 8 . If two triangles have three sides of PROPOSITION 7 . 29.
... ) ( 3 ) Next let the vertex D of one triangle lie on one of the sides BC of the other : then BC , BD are unequal . Wherefore , if two points on the same side & c . PROPOSITION 8 . If two triangles have three sides of PROPOSITION 7 . 29.
Side 34
... triangles are equal in all respects ; ( Prop . 4. ) therefore AD is equal to BD . Wherefore , the given finite straight line AB is bisected at the point D. EXERCISES . 1. Prove that there is only one point 34 BOOK 1 .
... triangles are equal in all respects ; ( Prop . 4. ) therefore AD is equal to BD . Wherefore , the given finite straight line AB is bisected at the point D. EXERCISES . 1. Prove that there is only one point 34 BOOK 1 .
Side 37
... at right angles to FG ; ( Prop . 10 A. ) therefore the angle A EC coincides with the angle FLH , and is equal to it . Wherefore , all right angles & c . PROPOSITION 11 . To draw a straight line at right PROPOSITION 10 B. 37.
... at right angles to FG ; ( Prop . 10 A. ) therefore the angle A EC coincides with the angle FLH , and is equal to it . Wherefore , all right angles & c . PROPOSITION 11 . To draw a straight line at right PROPOSITION 10 B. 37.
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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.