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 46
... fore AB.BC = AC.CB + BC2 . A C F D SCHOLIUM . B E In this proposition let AB be denoted by a , and the segments AC and CB , by b and c ; then a = b + c : therefore , multiplying both members of this equality by c , we shall have ac = bc ...
... fore AB.BC = AC.CB + BC2 . A C F D SCHOLIUM . B E In this proposition let AB be denoted by a , and the segments AC and CB , by b and c ; then a = b + c : therefore , multiplying both members of this equality by c , we shall have ac = bc ...
Página 108
... fore A is not greater than B ; and in the same way it is demonstrated that B is not greater than A ; therefore A is equal to B. Next , let C : A :: C : B , A = B . For by inversion ( A. 5. ) A : C :: B : C ; and therefore , by the first ...
... fore A is not greater than B ; and in the same way it is demonstrated that B is not greater than A ; therefore A is equal to B. Next , let C : A :: C : B , A = B . For by inversion ( A. 5. ) A : C :: B : C ; and therefore , by the first ...
Página 172
... fore the difference of either of them from the circumference must be less than the 497th part of the diameter . COR . 3. As 7 to 22 , so the square of the radius to the area of the circle nearly . For it has been shewn , that ( 1. Cor ...
... fore the difference of either of them from the circumference must be less than the 497th part of the diameter . COR . 3. As 7 to 22 , so the square of the radius to the area of the circle nearly . For it has been shewn , that ( 1. Cor ...
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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 demonstrated diameter divided draw Prob 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 solid parallelopiped spherical angle spherical triangle square straight line BC THEOR third touches the circle triangle ABC wherefore