Elements of Geometry; Containing the First Six Books of Euclid ...For Bell & Bradfute, 1814 - 461 Seiten |
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Seite 54
... half of the parallelogram EBCA , because the diameter AB bi- sects it ; and the triangle DBC is the half of the paral- lelogram DBCF , because the diameter DC bisects it : And the halves of equal things are equal ; therefore the ...
... half of the parallelogram EBCA , because the diameter AB bi- sects it ; and the triangle DBC is the half of the paral- lelogram DBCF , because the diameter DC bisects it : And the halves of equal things are equal ; therefore the ...
Seite 55
... half of the parallelogram GBCA , because the c54.1 . diameter AB bisects it ; and the triangle DEF is the half of the parallelogram DEFH , because the diameter DF bisects it : But the halves of equal things are equal d ; d 7. Ax ...
... half of the parallelogram GBCA , because the c54.1 . diameter AB bisects it ; and the triangle DEF is the half of the parallelogram DEFH , because the diameter DF bisects it : But the halves of equal things are equal d ; d 7. Ax ...
Seite 69
... half the line . Let the straight line AB be divided into two equal parts in the point C , and into two unequal parts in the point D ; the rectangle AD.DB , together with the square of CD , is equal to the square of CB , or AD.DB + CD2 ...
... half the line . Let the straight line AB be divided into two equal parts in the point C , and into two unequal parts in the point D ; the rectangle AD.DB , together with the square of CD , is equal to the square of CB , or AD.DB + CD2 ...
Seite 70
... half the line bisected , is equal to the square of the straight line which is made up of the half and the part produced . Let the straight line AB be bisected in C , and pro- duced to the point D ; the rectangle AD.DB , together with ...
... half the line bisected , is equal to the square of the straight line which is made up of the half and the part produced . Let the straight line AB be bisected in C , and pro- duced to the point D ; the rectangle AD.DB , together with ...
Seite 73
... half that line . " Otherwise : " Because AD is divided any how in C , AD2 = a 4. 2 . " AC2 + CD2 + 2CD.AC . But CD = 2CB : and there- " fore CD2 = CB2 + BD2 + 2CB.BD2 = 4CB2 , and also « 2CD.AC = 4CB.AC ; therefore , ADAC2 + 4BC2 " + ...
... half that line . " Otherwise : " Because AD is divided any how in C , AD2 = a 4. 2 . " AC2 + CD2 + 2CD.AC . But CD = 2CB : and there- " fore CD2 = CB2 + BD2 + 2CB.BD2 = 4CB2 , and also « 2CD.AC = 4CB.AC ; therefore , ADAC2 + 4BC2 " + ...
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ABC is equal ABCD altitude angle ABC angle ACB angle BAC arch AC base BC bisected Book centre circle ABC circumference cosine cylinder definition demonstrated described diameter draw drawn equal angles equiangular equilateral equilateral polygon equimultiples Euclid exterior angle fore four right angles GEOMETRY given straight line greater hypotenuse inscribed join less Let ABC Let the straight line BC magnitudes meet opposite angle parallel parallelogram perpendicular polygon prism PROB proportional 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 SPHERICAL TRIGONOMETRY straight line AC THEOR third touches the circle triangle ABC triangle DEF wherefore