Elements of Geometry; Containing the First Six Books of Euclid ...For Bell & Bradfute, 1814 - 461 Seiten |
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Seite 33
... perpendicular to a given straight line , of an unlimited length , from a gi- ven point without it . Let AB be a given straight line , which may be produ- ced to any length both ways , and let C be a point with- out it . It is required ...
... perpendicular to a given straight line , of an unlimited length , from a gi- ven point without it . Let AB be a given straight line , which may be produ- ced to any length both ways , and let C be a point with- out it . It is required ...
Seite 78
... perpendicular falls , and the straight line intercepted between the perpendicu- lar and the obtuse angle . Let ABC be an obtuse angled triangle , having the ob- tuse angle ACB , and from the point A let AD be drawn a perpendicular to BC ...
... perpendicular falls , and the straight line intercepted between the perpendicu- lar and the obtuse angle . Let ABC be an obtuse angled triangle , having the ob- tuse angle ACB , and from the point A let AD be drawn a perpendicular to BC ...
Seite 79
... perpendicular a AD from the opposite a 12.1 . angle : The square of AC , opposite to the angle B , is less than the squares of CB , BA by twice the rectangle CB.BD. First , Let AD fall within the triangle ABC ; and be- cause the ...
... perpendicular a AD from the opposite a 12.1 . angle : The square of AC , opposite to the angle B , is less than the squares of CB , BA by twice the rectangle CB.BD. First , Let AD fall within the triangle ABC ; and be- cause the ...
Seite 80
... perpendicular and the acute angle at B ; and it is manifest 47. 1. that AB2 + BC2 = AC2 + 2BC2 = AC2 + 2BC.BC . Therefore in every triangle , & c . Q. E. D. B a 45. 1 . b 5. 2 . € 47. 1 . PROP . XIV . PROB . To describe a square that ...
... perpendicular and the acute angle at B ; and it is manifest 47. 1. that AB2 + BC2 = AC2 + 2BC2 = AC2 + 2BC.BC . Therefore in every triangle , & c . Q. E. D. B a 45. 1 . b 5. 2 . € 47. 1 . PROP . XIV . PROB . To describe a square that ...
Seite 81
... perpendicular to BC , and because BEA is a right angle , AB a BE + AE and AC2 : CE + AE ; wherefore AB + AC BE + CE + 2AE * . But because the line BC is cut equally in D , and unequally in E , BE + CE = b2BD2 + 2DE2 ; therefore AB ' + ...
... perpendicular to BC , and because BEA is a right angle , AB a BE + AE and AC2 : CE + AE ; wherefore AB + AC BE + CE + 2AE * . But because the line BC is cut equally in D , and unequally in E , BE + CE = b2BD2 + 2DE2 ; therefore AB ' + ...
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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