Euclid's Elements of Geometry: The First Six, the Eleventh and Twelfth BooksJ. Rivington, 1765 - 464 sider |
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Resultat 1-5 av 76
Side 48
... thro ' K draw [ by prop . 31. ] KL , parallel to E A or F H. And continue out A H , G B to the points L , M. Then is HLKF a parallelogram , whofe diameter is HK , and the parallelograms about H K are A G , ME ; and these which are ...
... thro ' K draw [ by prop . 31. ] KL , parallel to E A or F H. And continue out A H , G B to the points L , M. Then is HLKF a parallelogram , whofe diameter is HK , and the parallelograms about H K are A G , ME ; and these which are ...
Side 75
... thro ' G [ by 31. 1. ] draw GH parallel to BC . And through the points D , E , c draw the right lines D K , EL , C H parallel to в G. F Then the rectangle B H is equal to the rectangles BK , DL , EH . And B H is the rectangle under A ...
... thro ' G [ by 31. 1. ] draw GH parallel to BC . And through the points D , E , c draw the right lines D K , EL , C H parallel to в G. F Then the rectangle B H is equal to the rectangles BK , DL , EH . And B H is the rectangle under A ...
Side 81
... thro ' the point B [ by 31. 1. ] draw BHG parallel to CE or DF ; through the point H draw K L M parallel to A D or EF ; and through a draw AK parallel to CL or D M. Now because A c is equal to CB , the rectangle A L will be [ by 36. 1 ...
... thro ' the point B [ by 31. 1. ] draw BHG parallel to CE or DF ; through the point H draw K L M parallel to A D or EF ; and through a draw AK parallel to CL or D M. Now because A c is equal to CB , the rectangle A L will be [ by 36. 1 ...
Side 102
... thro ' E draw GF parallel to the fide AB meeting the base A c in the point G , and the right line B F in the point F. B F 1 Now because [ by 15. 1. ] the angle B E F is equal to the angle GEC , and the angle EBF [ by 29. 1. ] is equal ...
... thro ' E draw GF parallel to the fide AB meeting the base A c in the point G , and the right line B F in the point F. B F 1 Now because [ by 15. 1. ] the angle B E F is equal to the angle GEC , and the angle EBF [ by 29. 1. ] is equal ...
Side 112
... thro ' the centre of a circle di- vides a right line not drawn thro ' the centre into two equal parts , it will cut the fame at right an- gles ; and if it cuts the fame at right angles , it will cut it into two equal parts . Let there ...
... thro ' the centre of a circle di- vides a right line not drawn thro ' the centre into two equal parts , it will cut the fame at right an- gles ; and if it cuts the fame at right angles , it will cut it into two equal parts . Let there ...
Andre utgaver - Vis alle
Euclid's Elements of Geometry: The First Six, the Eleventh and Twelfth Books Euclid,David Gregory Ingen forhåndsvisning tilgjengelig - 2023 |
Euclid's Elements of Geometry: The First Six, the Eleventh and Twelfth Books Euclid,David Gregory Ingen forhåndsvisning tilgjengelig - 2023 |
Euclid's Elements of Geometry: The First Six, the Eleventh and Twelfth Books Euclid,David Gregory Ingen forhåndsvisning tilgjengelig - 2016 |
Vanlige uttrykk og setninger
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Populære avsnitt
Side 247 - If two triangles have one angle of the one equal to one angle of the other and the sides about these equal angles proportional, the triangles are similar.
Side 30 - 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. either the sides adjacent to the equal...
Side 248 - But it was proved that the angle AGB is equal to the angle at F ; therefore the angle at F is greater than a right angle : But by the hypothesis, it is less than a right angle ; which is absurd.
Side 18 - When a straight line set up on a straight line makes the adjacent angles equal to one another, each of the equal angles is right, and the straight line standing on the other is called a perpendicular to that on which it stands.
Side 32 - Let the straight line EF, which falls upon the two straight lines AB, CD, make the alternate angles AEF, EFD equal to one another; AB is parallel to CD.
Side 56 - Therefore all the angles of the figure, together with four right angles, are equal to twice as many right angles as the figure has sides.
Side 391 - KL: but the cylinder CM is equal to the cylinder EB, and the axis LN to the axis GH; therefore as the cylinder EB to...
Side 110 - If any two points be taken in the circumference of a circle, the straight line which joins them shall fall within the circle.
Side 130 - When you have proved that the three angles of every triangle are equal to two right angles...
Side 183 - FK : in the same manner it may be demonstrated, that FL, FM, FG are each of them equal to FH, or FK : therefore the five straight lines FG, FH, FK, FL, FM are equal to one another : wherefore the circle described from the centre F, at the distance of...