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 45
... is a paral- E B A D lelogram ; and DBCA is equal ( 35. 1. ) to DBCF , because they are upon the same base BC , and between the same parallels BC , EF ; but the triangle ABC is the half of the parallelogram OF GEOMETRY . BOOK I. 45.
... is a paral- E B A D lelogram ; and DBCA is equal ( 35. 1. ) to DBCF , because they are upon the same base BC , and between the same parallels BC , EF ; but the triangle ABC is the half of the parallelogram OF GEOMETRY . BOOK I. 45.
Página 46
... half of the parallelogram EBCA , because the diameter AB bisects ( 34. 1. ) it ; and the triangle DBC is the half of the parallelogram DBCF , because the diameter DC bisects it ; and the halves of equal things are equal . ( 7. Ax ...
... half of the parallelogram EBCA , because the diameter AB bisects ( 34. 1. ) it ; and the triangle DBC is the half of the parallelogram DBCF , because the diameter DC bisects it ; and the halves of equal things are equal . ( 7. Ax ...
Página 56
... half the line . the Let the straight line AB be divided into two equal parts in the point C , and into two unequal ... half the line bisected , is equal to the square of the straight line which is made up of the half and the part ...
... half the line . the Let the straight line AB be divided into two equal parts in the point C , and into two unequal ... half the line bisected , is equal to the square of the straight line which is made up of the half and the part ...
Página 59
... half that line . " Otherwise : " Because AD is 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 ...
... half that line . " Otherwise : " Because AD is 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 ...
Página 60
... half and the part produced . " D ; Let the straight line AB be bisected in C , and produced to the point ; the ... half a right angle ( 32. 1. ) : For the same reason , each of the angles CEB , EBC is half a right angle ; therefore AEB ...
... half and the part produced . " D ; Let the straight line AB be bisected in C , and produced to the point ; the ... half a right angle ( 32. 1. ) : For the same reason , each of the angles CEB , EBC is half a right angle ; therefore AEB ...
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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