Solid and Spherical Geometry and Conic Sections: Being a Treatise on the Higher Branches of Synthetical Geometry, Containing the Solid and Spherical Geometry of Playfair ... |
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Resultat 1-5 av 43
Side 3
In OD take any point G , and make GD = OG , and through G draw GF parallel to
OB , to meet OF in F ; join DF , and produce DF to meet AB in B , and join PD , PF
, and PB . Because OG = GD , therefore ( Pl . Ge . VI . 2 . ) BF = FD ; and because
...
In OD take any point G , and make GD = OG , and through G draw GF parallel to
OB , to meet OF in F ; join DF , and produce DF to meet AB in B , and join PD , PF
, and PB . Because OG = GD , therefore ( Pl . Ge . VI . 2 . ) BF = FD ; and because
...
Side 5
Let them meet the plane in the points B , D , and draw the straight line BD , to
which draw DE at right angles , in the same plane ; and make DE equal to AB ,
and join BE , AE , AD . Then , because AB Ak 35 , 131c is perpendicular to the
plane ...
Let them meet the plane in the points B , D , and draw the straight line BD , to
which draw DE at right angles , in the same plane ; and make DE equal to AB ,
and join BE , AE , AD . Then , because AB Ak 35 , 131c is perpendicular to the
plane ...
Side 7
Take BA , BC , ED , EF , all equal to one another ; and join AD , CF , BE , A AC ,
DF ; because BA is equal and parallel to ED , therefore AD is ( Pl . Ge . I . 33 )
both equal and parallel to BE ; for the same reason , CF is equal and parallel to
BE ...
Take BA , BC , ED , EF , all equal to one another ; and join AD , CF , BE , A AC ,
DF ; because BA is equal and parallel to ED , therefore AD is ( Pl . Ge . I . 33 )
both equal and parallel to BE ; for the same reason , CF is equal and parallel to
BE ...
Side 9
If not , they shall meet one another when produced ; let them meet ; their common
section shall be a straight line GH , in which take any point K , and join AK , BK ;
then , because AB is perpendicular to the plane EF , it is perpendicular ( I . Def .
If not , they shall meet one another when produced ; let them meet ; their common
section shall be a straight line GH , in which take any point K , and join AK , BK ;
then , because AB is perpendicular to the plane EF , it is perpendicular ( I . Def .
Side 11
Join AC , BD , AD , and let AD meet the plane KL in the point X ; and join EX , XF .
Because the two parallel planes KL , MN , are cut by the plane EBDX , G the
common sections EX , BD , are parallel LR Ja ( I . 3 ) . For the same reason ...
Join AC , BD , AD , and let AD meet the plane KL in the point X ; and join EX , XF .
Because the two parallel planes KL , MN , are cut by the plane EBDX , G the
common sections EX , BD , are parallel LR Ja ( I . 3 ) . For the same reason ...
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Vanlige uttrykk og setninger
ABCD affection altitude angle ABC axis base bisects called centre circle common section cone conjugate consequently contained cord cosine curve cylinder described diameter difference distance divided draw drawn ellipse equal extremities fall figure foci focus fore given given point greater half Hence hyperbola inclination intercepted intersection join less line be drawn lines drawn manner measure meet namely opposite ordinate parabola parallel parallelogram pass perpendicular perspective plane point of contact pole primitive prism produced projection proportional PROPOSITION proved pyramid quadrant radius ratio reason rectangle right angles segments semi-ordinate sides similar sine small circle solid sphere spherical triangle square straight line surface tangent THEOREM third transverse triangle vertex vertical
Populære avsnitt
Side 50 - If two triangles have two angles of the one equal to two angles of the other, each to each, and one side equal to one side, viz.
Side 15 - A cone is a solid figure described by the revolution of a right-angled triangle about one of the sides containing the right angle, which side remains fixed.
Side 25 - LR, the base of which is the parallelogram LQ, and of which LM is one of its insisting straight lines : therefore, because the parallelogram AB is equal to CD, as the base AB is to the base LQ, so is (7.
Side 17 - DAB, which contain the solid angle at A, are less than four right angles. Next, let the solid angle at A be contained by any number of plane angles BAC, CAD, DAE, EAF, FAB. These shall together be less than four right angles.
Side 27 - FC, as the solid HD to the solid DC. But the base HF is equal to the base AE, and the solid GK to the solid AB ; therefore, as the base AE to the base CF, so is the solid AB to the solid CD.
Side 53 - EM (2.) are ^quadrants, and FL, EM together, that is, FE and ML together, are equal to a semicircle. But since A is the pole of ML, ML is the measure of the angle BAC (3.), consequently FE is the supplement of the measure of the angle BAC.
Side 19 - And AB is parallel to CD ; therefore AC is a parallelogram. In like manner, it may be proved, that each of the figures CE, FG, GB, BF, AE is a parallelogram: Join AH, DF; and...
Side 5 - If two straight lines be at right angles to the same plane, they shall be parallel to one another. Let the straight lines AB, CD be at right angles to the same plane.
Side 9 - CA is at right angles to the given plane, it makes right angles with every straight line meeting it in that plane. But DAE, which is in that plane, meets CA : therefore CAE is a right angle. For the same reason BAE is a right angle. Wherefore the angle CAE is equal to the angle BAE ; and they are in one plane, which is impossible. Also, from a point above a plane, there can be but one perpendicular to that plane ; for if there could be two, they would be parallel (6.
Side 1 - The inclination of a straight line to a plane is the acute angle contained by that straight line, and another drawn from the point in which the first line meets the plane, to the point in which...