The Elements of EuclidJohnson and Warner, 1811 - 518 sider |
Inni boken
Resultat 6-10 av 91
Side 82
... perpendicular to FA , and let it meet the circumference in H : and because AGD is a right angle , and EAG less than a right angle : DA is greater than DG : but DA is equal to DH ; therefore DH is greater than DG , the less than the ...
... perpendicular to FA , and let it meet the circumference in H : and because AGD is a right angle , and EAG less than a right angle : DA is greater than DG : but DA is equal to DH ; therefore DH is greater than DG , the less than the ...
Side 83
... perpendicular to the line touching the circle . Let the straight line DE touch the circle ABC in the point C ; take the centre F , and draw the straight line FC : FC is perpendicular to DE . For , if it be not , from the point F draw ...
... perpendicular to the line touching the circle . Let the straight line DE touch the circle ABC in the point C ; take the centre F , and draw the straight line FC : FC is perpendicular to DE . For , if it be not , from the point F draw ...
Side 84
... perpendicular to it besides FC , that is FC is perpendicular to DE . Therefore , if a straight line , & c . Q. E. D. 18.3 . D 30 the yo PROP . XIX . THEOR . A F E IF a straight line touch a circle , and from the point of contact a ...
... perpendicular to it besides FC , that is FC is perpendicular to DE . Therefore , if a straight line , & c . Q. E. D. 18.3 . D 30 the yo PROP . XIX . THEOR . A F E IF a straight line touch a circle , and from the point of contact a ...
Side 98
... perpendicular to AC ; therefore AG is equal to GC ; where- fore the rectangle AE , EC , together with the square of EG , is equal to the square of AG : to each of these equals add the square of GF ; therefore the rectangle AE , EC ...
... perpendicular to AC ; therefore AG is equal to GC ; where- fore the rectangle AE , EC , together with the square of EG , is equal to the square of AG : to each of these equals add the square of GF ; therefore the rectangle AE , EC ...
Side 99
... perpendicular e tod 1. 3 . AC , and join EB , EC , ED : and because the straight line EF , e 12. 1 . which passes through the centre , cuts the straight line AC , which does not pass through the centre , at right angles , it shall ...
... perpendicular e tod 1. 3 . AC , and join EB , EC , ED : and because the straight line EF , e 12. 1 . which passes through the centre , cuts the straight line AC , which does not pass through the centre , at right angles , it shall ...
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The Elements of Euclid: The Errors, by which Theon, Or Others, Have Long Ago ... Robert Simson Uten tilgangsbegrensning - 1806 |
Vanlige uttrykk og setninger
altitude angle ABC angle BAC base BC BC is equal BC is given bisected Book XI centre circle ABCD circumference cone cylinder demonstrated described diameter draw drawn equal angles equiangular equimultiples Euclid excess fore given angle given in magnitude given in position given in species given magnitude given ratio given straight line gnomon greater join less Let ABC meet multiple parallel parallelogram parallelogram AC perpendicular point F polygon prism proportionals proposition pyramid Q. E. D. PROP radius ratio of AE rectangle CB rectangle contained rectilineal figure remaining angle right angles segment sides BA similar solid angle solid parallelopiped square of AC straight line AB straight line BC tangent THEOR third triangle ABC triplicate ratio vertex wherefore