Elements of Geometry: Containing the First Six Books of Euclid, with a Supplement on the Quadrature of the Circle, and the Geometry of Solids ; to which are Added, Elements of Plane and Spherical TrigonometryW.E. Dean, 1836 - 311 páginas |
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Página 92
... square ABCD . PROP . IX . PROB . To divide a given straight line into two parts , so that the rectangle contained by the whole , and one of the parts , may be equal to the square of the other part . Let AB be the given straight line ...
... square ABCD . PROP . IX . PROB . To divide a given straight line into two parts , so that the rectangle contained by the whole , and one of the parts , may be equal to the square of the other part . Let AB be the given straight line ...
Página 149
... square of EG , is equal to the square of GH : But the squares of HE and EG are equal ( 37. 1. ) to the square of GH : Therefore also the rectangle BE.EF , together with the square of EG , is equal to the squares of HE and EG . Take away the ...
... square of EG , is equal to the square of GH : But the squares of HE and EG are equal ( 37. 1. ) to the square of GH : Therefore also the rectangle BE.EF , together with the square of EG , is equal to the squares of HE and EG . Take away the ...
Página 162
... square of HG to the square of AG , that is of GK . But the triangles HGK , ACF have been proved to be similar , and therefore the square of AC is to the square of CF as the polygon M to the polygon N ; and , by conversion , the ...
... square of HG to the square of AG , that is of GK . But the triangles HGK , ACF have been proved to be similar , and therefore the square of AC is to the square of CF as the polygon M to the polygon N ; and , by conversion , the ...
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Términos y frases comunes
ABC is equal ABCD adjacent angles altitude angle ABC angle ACB angle BAC angles equal arc AC base BC bisected centre chord circle ABC circumference cosine cylinder definition demonstrated diameter divided equal and similar equal angles equiangular equilateral equilateral polygon equimultiples Euclid exterior angle fore four right angles given straight line greater Hence hypotenuse inscribed join less Let ABC Let the straight magnitudes meet multiple opposite angle parallel parallelogram parallelopiped perpendicular polygon prism PROP proposition quadrilateral radius ratio rectangle contained rectilineal figure remaining angle right angled triangle SCHOLIUM segment semicircle shewn side BC sine solid angle spherical angle spherical triangle square straight line BC THEOR third touches the circle triangle ABC triangle DEF wherefore