A Supplement to the Elements of EuclidJ. Deighton and Sons, Cambridge and G and W. B. Whittaker, Ave-Maria-Lane, 1819 - 410 σελίδες |
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Σελίδα 5
... COR . 4. Hence , also , a circle may be de- scribed which shall pass through two given points , and which shall have its semi - diameter equal to any given finite straight line , that exceeds the half ELEMENTS OF EUCLID . 5.
... COR . 4. Hence , also , a circle may be de- scribed which shall pass through two given points , and which shall have its semi - diameter equal to any given finite straight line , that exceeds the half ELEMENTS OF EUCLID . 5.
Σελίδα 6
... diameter , will pass through B. PROP . IV . 8. THEOREM . If the three sides of a given triangle be bisected , the perpendiculars drawn to the sides , from the three several bisections , shall all meet in the same point : And that point ...
... diameter , will pass through B. PROP . IV . 8. THEOREM . If the three sides of a given triangle be bisected , the perpendiculars drawn to the sides , from the three several bisections , shall all meet in the same point : And that point ...
Σελίδα 35
... diameter , to describe a square . Let AB be a given finite straight line : Upon D E AB , as a diameter , it is required to describe a square . Bisect ( E. 10. 1. ) AB in E ; through E draw ( E. 11. 1. ) DECT to AB , and make ( E. 3 ...
... diameter , to describe a square . Let AB be a given finite straight line : Upon D E AB , as a diameter , it is required to describe a square . Bisect ( E. 10. 1. ) AB in E ; through E draw ( E. 11. 1. ) DECT to AB , and make ( E. 3 ...
Σελίδα 52
... diameter of ABCD , and FH a diameter of EFGH ; let AC and FH cut one an- other in K ; and let CB , produced , meet EF , pro- duced , in L : Then , since AE is parallel to BC , and EF parallel to HG , the CGH = ( E . 29. 1. ) = GLE ...
... diameter of ABCD , and FH a diameter of EFGH ; let AC and FH cut one an- other in K ; and let CB , produced , meet EF , pro- duced , in L : Then , since AE is parallel to BC , and EF parallel to HG , the CGH = ( E . 29. 1. ) = GLE ...
Σελίδα 53
... diameters AC , FH ; ( S. 42. 1. ) all the diameters cut one another in the point K. 60. COR . From the demonstration it is mani- fest , that the angle contained by any two given straight lines , is equal to the angle contained by two ...
... diameters AC , FH ; ( S. 42. 1. ) all the diameters cut one another in the point K. 60. COR . From the demonstration it is mani- fest , that the angle contained by any two given straight lines , is equal to the angle contained by two ...
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ABCD 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 point G 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 tangent THEOREM three given touch the circle trapezium vertex