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 TrigonometryB. & S. Collins; W. E. Dean, printer, 1836 - 311 páginas |
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Página 46
... being a right angle , the other angles of the parallelogram CGKB are also right angles Wherefore CGKB is a square , and it is upon the side CB . D F E ( Cor . 2. 28. 1. ) For the same reason HF also is a square , and it is 46 ELEMENTS.
... being a right angle , the other angles of the parallelogram CGKB are also right angles Wherefore CGKB is a square , and it is upon the side CB . D F E ( Cor . 2. 28. 1. ) For the same reason HF also is a square , and it is 46 ELEMENTS.
Página 47
... reason HF also is a square , and it is upon the side HG , which is equal to AC : therefore HF , CK are the squares of AC , CB . And because the complement AG is equal ( 36. 1. ) to the complement GE ; and because AG = AC.CG AC.CB ...
... reason HF also is a square , and it is upon the side HG , which is equal to AC : therefore HF , CK are the squares of AC , CB . And because the complement AG is equal ( 36. 1. ) to the complement GE ; and because AG = AC.CG AC.CB ...
Página 49
... reason , PR is equal to RO ; and because CB is equal to BD , and GK to KN , the rectangles CK and BN are equal , as also the rectangles GR and RN : But CK is equal ( 36. 1. ) to RN , because they are the complements of the parallelogram ...
... reason , PR is equal to RO ; and because CB is equal to BD , and GK to KN , the rectangles CK and BN are equal , as also the rectangles GR and RN : But CK is equal ( 36. 1. ) to RN , because they are the complements of the parallelogram ...
Página 50
... together make one right angle ( Cor . 4. 25. 1. ) ; and they are equal to one ano- ther ; each of them therefore is half of a right angle . For the same reason each J A C DO B of the angles CEB , EBC is half a right 50 ELEMENTS.
... together make one right angle ( Cor . 4. 25. 1. ) ; and they are equal to one ano- ther ; each of them therefore is half of a right angle . For the same reason each J A C DO B of the angles CEB , EBC is half a right 50 ELEMENTS.
Página 54
... reason , CD2 + BC2 = 2BE2 + 2EC2 = 2BE2 + 2AE2 , because ECAE . Therefore AB2 + AD2 + DC2 + BC2 = 4BE2 + 4AE2 . But 4BE2 = BD2 , and 4AE2 - AC2 ( 2 Cor . 8. 2. ) because BD and AC are both bisected in E ; therefore AB2 + AC2 + CD2 + BC2 ...
... reason , CD2 + BC2 = 2BE2 + 2EC2 = 2BE2 + 2AE2 , because ECAE . Therefore AB2 + AD2 + DC2 + BC2 = 4BE2 + 4AE2 . But 4BE2 = BD2 , and 4AE2 - AC2 ( 2 Cor . 8. 2. ) because BD and AC are both bisected in E ; therefore AB2 + AC2 + CD2 + BC2 ...
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ABC is equal ABCD adjacent angles altitude angle ABC angle ACB angle BAC angles equal base BC bisected centre chord circle ABC circumference cosine cylinder 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 solid parallelopiped spherical angle spherical triangle square straight line BC THEOR third touches the circle triangle ABC triangle DEF wherefore