A Supplement to the Elements of EuclidG. and W. B. Whittaker, 1819 - 410 sider |
Inni boken
Resultat 6-10 av 21
Side 33
... falls within EB ; the CAE is a right , the than a right ; .. ( E. 17. 1. ) the than a right , and .. less than the DAE ; .. ( E. 19. 1. ) DE > DA ; but , by the former case , DA = DB ; ... DE > DB , or DC . DAE is greater AED is less L ...
... falls within EB ; the CAE is a right , the than a right ; .. ( E. 17. 1. ) the than a right , and .. less than the DAE ; .. ( E. 19. 1. ) DE > DA ; but , by the former case , DA = DB ; ... DE > DB , or DC . DAE is greater AED is less L ...
Side 46
... fall from the extremities of the two so drawn , make , with the base pro . duced , two angles that are equal each of them to the vertical angle , they shall cut off equal seg- ments from the base produced . From the extremities B , C ...
... fall from the extremities of the two so drawn , make , with the base pro . duced , two angles that are equal each of them to the vertical angle , they shall cut off equal seg- ments from the base produced . From the extremities B , C ...
Side 69
... falls without the given figure : In FE take any point E , and join E , A , and E , D : The figure ABCDE has more sides , by one , than the given figure ABCF ; and it may be shewn , as in the preceding case , to be equal to ABCF . 75 ...
... falls without the given figure : In FE take any point E , and join E , A , and E , D : The figure ABCDE has more sides , by one , than the given figure ABCF ; and it may be shewn , as in the preceding case , to be equal to ABCF . 75 ...
Side 101
... fall within the base BC . Then , BD ' + DC = 2DE + 2EB . ( E. 9. 2. ) Add to these equals 2AD ' . ... BD2 + DC + 2AD * = 2AD ' + 2DE * + 2EB * • i . e . AB ' + AC2 = 2AE + 2EB . ( E. 47. 1. ) And , if the perpendicular AD fall without ...
... fall within the base BC . Then , BD ' + DC = 2DE + 2EB . ( E. 9. 2. ) Add to these equals 2AD ' . ... BD2 + DC + 2AD * = 2AD ' + 2DE * + 2EB * • i . e . AB ' + AC2 = 2AE + 2EB . ( E. 47. 1. ) And , if the perpendicular AD fall without ...
Side 103
... fall without AD , so that both of them meet AD produced . .. ACDC + AD2 + 2AD × DF ) and BD ' = AB + AD2 + 2AD × AE ( E. 12. 2. ) .. AC + BD = AB ' + DC ' + 2AD ' + 2AD XAE × + 2ADXDF . But 2AD * + 2ADXAE + 2AD × DF = ELEMENTS OF EUCLID ...
... fall without AD , so that both of them meet AD produced . .. ACDC + AD2 + 2AD × DF ) and BD ' = AB + AD2 + 2AD × AE ( E. 12. 2. ) .. AC + BD = AB ' + DC ' + 2AD ' + 2AD XAE × + 2ADXDF . But 2AD * + 2ADXAE + 2AD × DF = ELEMENTS OF EUCLID ...
Andre utgaver - Vis alle
Vanlige uttrykk og setninger
base BC bisect centre chord circle ABC circle described circumference constr decagon describe a circle describe the circle diameter distance divided double draw a straight draw E equi equiangular equilateral and equiangular F draw find a point finite straight line given circle given finite straight given point given ratio given square given straight line half hypotenuse inscribed isosceles less Let AB Let ABC lines be drawn magnitudes manifest manner meet the circumference number of equal number of sides parallel to BC parallelogram pass perimeter polygon PROBLEM produced PROP rectangle contained rectilineal figure remaining sides required to describe required to draw rhombus right angles segment semi-diameter straight line joining subtend tangent THEOREM three given touch the circle trapezium vertex
Populære avsnitt
Side 310 - If two triangles have one angle of the one equal to one angle of the other and the sides about these equal angles proportional, the triangles are similar.
Side 198 - 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, together with the square...
Side 366 - If from the vertical angle of a triangle a straight line be drawn perpendicular to the base, the rectangle contained by the sides of the triangle is equal to the rectangle contained by the perpendicular and the diameter of the circle described about the...
Side 92 - 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 284 - And if the first have a greater ratio to the second, than the third has to the fourth, but the third the same ratio to the fourth, which the fifth has to the sixth...
Side 349 - Divide a straight line into two parts such that the rectangle contained by the whole line and one of the parts shall be equal to the square on the other part.
Side 288 - Convertendo, by conversion ; when there are four proportionals, and it is inferred, that the first is to its excess above the second, as the third to its excess above the fourth.
Side 296 - ... line and the extremities of the base have the same ratio which the other sides of the triangle have to one...
Side 367 - Wherefore, in equal circles &c. QED PROPOSITION B. THEOREM If the vertical angle of a triangle be bisected by a straight line which likewise cuts the base, the rectangle contained by the sides of the triangle is equal to the rectangle contained by the segments of the base, together with the square on the straight line which bisects the angle.
Side 104 - In every triangle, the square of the side subtending any of the acute angles is less than the squares of the sides containing that angle by twice the rectangle contained by either of these sides, and the straight line intercepted between the perpendicular let fall upon it from the opposite angle, and the acute angle. Let ABC be any triangle, and the angle at B one of its acute angles, and upon BC, one of the sides containing it, let fall the perpendicular...