Elements of Geometry |
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Page 5
... drawn from A to B , which is impossi- ble ( 25 ) . Now let us suppose , if it be possible , that the lines , when produced , separate from each other at a point C , the one becoming CD , and the other CE . At the point C , let CF be . drawn ...
... drawn from A to B , which is impossi- ble ( 25 ) . Now let us suppose , if it be possible , that the lines , when produced , separate from each other at a point C , the one becoming CD , and the other CE . At the point C , let CF be . drawn ...
Page 9
... Draw the straight line AD from the vertex A to the point D the middle of the base BC ; the two triangles ABD , ADC , will have the three sides of the one , equal to the three sides of the other , each to each , namely , AD common to ...
... Draw the straight line AD from the vertex A to the point D the middle of the base BC ; the two triangles ABD , ADC , will have the three sides of the one , equal to the three sides of the other , each to each , namely , AD common to ...
Page 10
... drawn from the vertex of an isosceles triangle , to the middle of the base , is perpendicular to the base , and ... Draw CD making the angle BCD = B . In the triangle BDC , BD is equal to DC ( 48 ) ; but AD + DC > AC , and AD + DC ...
... drawn from the vertex of an isosceles triangle , to the middle of the base , is perpendicular to the base , and ... Draw CD making the angle BCD = B . In the triangle BDC , BD is equal to DC ( 48 ) ; but AD + DC > AC , and AD + DC ...
Page 11
... drawn to that line . Demonstration . If it be possible , let there be two AB and AC ; produce one of them AB , so ... draw two perpendiculars to that line ; for , if CD and CE were these two perpendiculars , the an gle DCB would be a ...
... drawn to that line . Demonstration . If it be possible , let there be two AB and AC ; produce one of them AB , so ... draw two perpendiculars to that line ; for , if CD and CE were these two perpendiculars , the an gle DCB would be a ...
Page 12
... drawn three equal straight lines terminating in a given straight line ; for , if this could be done , there would be on the same side of the perpendicular two equal oblique lines , which is impossible . Fig . 32 THEOREM . 55. If from ...
... drawn three equal straight lines terminating in a given straight line ; for , if this could be done , there would be on the same side of the perpendicular two equal oblique lines , which is impossible . Fig . 32 THEOREM . 55. If from ...
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Common terms and phrases
ABC fig adjacent angles altitude angle ACB angle BAD base ABCD bisect centre chord circ circular sector circumference circumscribed common cone consequently construction convex surface Corollary cube cylinder Demonstration diagonals diameter draw drawn equal and parallel equiangular equilateral equivalent faces figure four right angles frustum Geom gles greater hence homologous sides hypothenuse inclination inscribed circle isosceles join less let fall line AC manner mean proportional measure the half meet multiplied number of sides oblique lines opposite parallelogram parallelopiped perimeter perpendicular plane MN polyedron prism proposition pyramid S-ABC quadrilateral radii radius ratio rectangle regular polygon right angles right-angled triangle Scholium segment semicircumference side BC similar solid angle sphere spherical polygons spherical triangle square described straight line tangent THEOREM three plane angles triangle ABC triangular prism triangular pyramids vertex vertices whence