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 TrigonometryE. Duyckinck, and George Long, 1824 - 333 páginas |
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Página 56
... divided & c . Q. E. D. COR . From the demonstration , it is manifest that the parallelograms about the diameter of a square are likewise squares . PROP . V. THEOR . If a straight line be divided into two equal parts , and also into two ...
... divided & c . Q. E. D. COR . From the demonstration , it is manifest that the parallelograms about the diameter of a square are likewise squares . PROP . V. THEOR . If a straight line be divided into two equal parts , and also into two ...
Página 57
... divided into any two parts , the squares of the whole line , and of one of the parts , are equal to twice the rectangle con tained by the whole and that part , together with the square of the other part . A C B K Let the straight line ...
... divided into any two parts , the squares of the whole line , and of one of the parts , are equal to twice the rectangle con tained by the whole and that part , together with the square of the other part . A C B K Let the straight line ...
Página 58
... divided into any two parts , four times the rectangle contained by the whole line , and one of the parts , together with the square of the other part , is equal to the square of the straight line which is made up of the whole and the ...
... divided into any two parts , four times the rectangle contained by the whole line , and one of the parts , together with the square of the other part , is equal to the square of the straight line which is made up of the whole and the ...
Página 59
... divided any how in C ( 4. 2. ) , AD2 = AC2 + CD2 " + 2CD.AC . But CD = 2CB and therefore CD CB2 + BD2 + " 2CB.BD ( 4. 2 . ) = 4CB2 , and also 2CD.AC-4CB.AC ; therefore , " AD2 = AC2 + 4BC2 + 4BC.AC . Now BC2 + BC.AC = AB.BC ( 3.2 ...
... divided any how in C ( 4. 2. ) , AD2 = AC2 + CD2 " + 2CD.AC . But CD = 2CB and therefore CD CB2 + BD2 + " 2CB.BD ( 4. 2 . ) = 4CB2 , and also 2CD.AC-4CB.AC ; therefore , " AD2 = AC2 + 4BC2 + 4BC.AC . Now BC2 + BC.AC = AB.BC ( 3.2 ...
Página 61
... divided in H , so that the rectangle AB , BH is equal to the square of AH . F G Produce GH to K : Because the straight line AC is bisected in E , and produced to the point F , the rectangle CF.FA , together with the square of AE , is ...
... divided in H , so that the rectangle AB , BH is equal to the square of AH . F G Produce GH to K : Because the straight line AC is bisected in E , and produced to the point F , the rectangle CF.FA , together with the square of AE , is ...
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Términos y frases comunes
ABC is equal ABCD altitude angle ABC angle ACB angle BAC angle EDF arch AC base BC bisected centre circle ABC circumference cosine cylinder demonstrated diameter draw equal angles equiangular equilateral equilateral polygon equimultiples Euclid exterior angle fore four right angles given straight line greater hypotenuse inscribed join less Let ABC Let the straight line BC magnitudes meet multiple opposite angle parallel parallelepipeds parallelogram perpendicular polygon prism PROB produced proportionals proposition Q. E. D. COR Q. E. D. PROP radius ratio rectangle contained rectilineal figure remaining angle segment semicircle shewn side BC sine solid angle solid parallelepipeds spherical angle spherical triangle straight line AC THEOR third touches the circle triangle ABC triangle DEF wherefore