Stewart's specific subjects. Euclid. [1st] (-3rd stage). [With 2 issues of] Algebra |
Inni boken
Resultat 1-5 av 19
Side 10
... bisect a given rectilineal angle , that is , divide it into two equal angles . Let BAC be the given rectilineal angle , it is re- quired to bisect it . Take any point D in AB , and from AC cut off AE , equal to AD ; join DE , and on it ...
... bisect a given rectilineal angle , that is , divide it into two equal angles . Let BAC be the given rectilineal angle , it is re- quired to bisect it . Take any point D in AB , and from AC cut off AE , equal to AD ; join DE , and on it ...
Side 12
... bisect FG in H , and join CF , CH , CG ; the straight line CH , drawn from the given point C , is perpendicular to the given straight line AB . D G B Because FH is equal to HG , and HC common to the two triangles FHC , GHC , the two ...
... bisect FG in H , and join CF , CH , CG ; the straight line CH , drawn from the given point C , is perpendicular to the given straight line AB . D G B Because FH is equal to HG , and HC common to the two triangles FHC , GHC , the two ...
Side 4
... Bisect AC in E , join BE and produce it to F , and make EF equal to BE ; join also FC , and produce AC to G. Because AE is equal to EC , and BE to EF ; AE , EB , are equal to CE , EF , each to each ; and the angle AEB is equal to CEF ...
... Bisect AC in E , join BE and produce it to F , and make EF equal to BE ; join also FC , and produce AC to G. Because AE is equal to EC , and BE to EF ; AE , EB , are equal to CE , EF , each to each ; and the angle AEB is equal to CEF ...
Side 4
... Bisect AC in E , join BE and produce it to F , and make EF equal to BE ; join also FC , and produce AC to G. E Because AE is equal to EC , and BE to EF ; AE , EB , are equal to CE , EF , each to each ; and the angle AEB is equal to CEF ...
... Bisect AC in E , join BE and produce it to F , and make EF equal to BE ; join also FC , and produce AC to G. E Because AE is equal to EC , and BE to EF ; AE , EB , are equal to CE , EF , each to each ; and the angle AEB is equal to CEF ...
Side 15
... bisects them . Let ACDB be a parallelogram , of which BC is a diameter . Because AB is parallel to CD , and BC meets them , the alternate angles ABC , BCD , are equal to one another ; and A ACB , CBD , are equal to one another ...
... bisects them . Let ACDB be a parallelogram , of which BC is a diameter . Because AB is parallel to CD , and BC meets them , the alternate angles ABC , BCD , are equal to one another ; and A ACB , CBD , are equal to one another ...
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Stewart's Specific Subjects. Euclid. [1st] (-3rd Stage). [With 2 Issues Of ... Stewart W and Co Ingen forhåndsvisning tilgjengelig - 2015 |
Vanlige uttrykk og setninger
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Populære avsnitt
Side 19 - If two triangles have two sides of the one equal to two sides of the...
Side 1 - When a straight line standing on another straight line makes the adjacent angles equal to one another, each of the angles is called a Right Angle; and the straight line which stands on the other is called a Perpendicular to it.
Side 8 - Upon the same base, and on the same side of it, there cannot be two triangles, that have their sides which are terminated in one extremity of the base equal to one another, and likewise those which are terminated in the other extremity, equal to one another.
Side 16 - Any two sides of a triangle are together greater than the third side.
Side 12 - IF a straight line be divided into any two parts, four times the rectangle contained by the whole line, and one of the parts, together with the square of the other part, is equal to the square of the straight line which is made up of the whole and that part.
Side 5 - IF two angles of a triangle be equal to one another, the sides also which subtend, or are opposite to, the equal angles, shall be equal to one another.