The Elements of Euclid, with many additional propositions, and explanatory notes, by H. Law. Pt. 2, containing the 4th, 5th, 6th, 11th, & 12th books1855 |
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Side 6
... both circles . CONSTRUCTION . Bisect the angles BCD and CDE ( a ) , by the straight lines CF and DF , then because the angles FCD and FDC H ( a ) I. 9 . E are together less than two right angles , there- fore 6 ELEMENTS OF GEOMETRY .
... both circles . CONSTRUCTION . Bisect the angles BCD and CDE ( a ) , by the straight lines CF and DF , then because the angles FCD and FDC H ( a ) I. 9 . E are together less than two right angles , there- fore 6 ELEMENTS OF GEOMETRY .
Side 7
Euclides Henry Law. are together less than two right angles , there- fore CF and DF will meet , if produced far enough ( b ) , let them meet in F. Join BF , and from F draw GF and HF respectively B perpendicular to BC and CD ( c ) . H E ...
Euclides Henry Law. are together less than two right angles , there- fore CF and DF will meet , if produced far enough ( b ) , let them meet in F. Join BF , and from F draw GF and HF respectively B perpendicular to BC and CD ( c ) . H E ...
Side 10
... fore the angles BDA and B are each double of the angle A. I. 6 . COROLLARY 1. The triangle BDC is also isosceles , and has each of the angles at its base , B and BCD , double of the vertical angle BDC . COROLLARY 2. The triangle ACD is ...
... fore the angles BDA and B are each double of the angle A. I. 6 . COROLLARY 1. The triangle BDC is also isosceles , and has each of the angles at its base , B and BCD , double of the vertical angle BDC . COROLLARY 2. The triangle ACD is ...
Side 13
... fore the side AF is equal to FE ( 6 ) ; and in the same manner it may be shown that all the lines AF , BF , CF , DF , and EF are equal , and therefore the circle described from F as a center with the radius AF will pass through the ...
... fore the side AF is equal to FE ( 6 ) ; and in the same manner it may be shown that all the lines AF , BF , CF , DF , and EF are equal , and therefore the circle described from F as a center with the radius AF will pass through the ...
Side 15
... fore the arc BC contains two , and therefore the arc BE is the fifteenth part of the whole circumference , and BE is the side of the required equilateral and equiangular quindecagon . SCHOLIUM . The only regular polygons which the Greek ...
... fore the arc BC contains two , and therefore the arc BE is the fifteenth part of the whole circumference , and BE is the side of the required equilateral and equiangular quindecagon . SCHOLIUM . The only regular polygons which the Greek ...
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Vanlige uttrykk og setninger
algebraically expressed altitude angle ABC angle BAC axis base ABC base DEF base EH circle ABCD circle EFGH circumference common section cone contained COROLLARY cylinder DEMONSTRATION diameter divided duplicate ratio equal and similar equal angles equi equiangular equimultiples Euclid ex æquali fore four magnitudes fourth given circle given straight line gnomon greater ratio homologous sides Hypoth inscribed join less meet multiple opposite planes paral parallel parallelogram pentagon perpendicular polygon prism PROPOSITION pyramid ABCG pyramid DEFH reciprocally proportional rectangle rectilineal figure remaining angle right angles SCHOLIUM segments solid angle solid CD solid parallelopipeds solid polyhedron square on BD THEOREM THEOREM.-If third three plane angles tiple triangle ABC triplicate ratio vertex vertex the point wherefore
Populære avsnitt
Side 198 - ... have an angle of the one equal to an angle of the other, and the sides about those angles reciprocally proportional, are equal to une another.
Side 75 - ... if the segments of the base have the same ratio which the other sides of the triangle have to one another...
Side 115 - If two triangles have two angles of the one equal to two angles of the other, each to each, and also one side of the one equal to the corresponding side of the other, the triangles are congruent.
Side 82 - From the point A draw a straight line AC, making any angle with AB ; and in AC take any point D, and take AC the same multiple of AD, that AB is of the part which is to be cut off from it : join BC, and draw DE parallel to it : then AE is the part required to be cut off.
Side 198 - Equiangular parallelograms have to one another the ratio which is compounded of the ratios of their sides.
Side 53 - Convertendo, by conversion ; when there are four proportionals, and it is inferred, that the first is to its excess above the second, as the third to its excess above the fourth.
Side 40 - A and B are not unequal ; that is, they are equal. Next, let C have the same ratio to each of the magnitudes A and B ; then A shall be equal to B.
Side 119 - For the same reason, CD is likewise at right angles to the plane HGK. Therefore AB, CD are each of them at right angles to the plane HGK.
Side 115 - FB ; (i. 4.) for the same reason, CF is equal to FD : and because AD is equal to BC, and AF to FB, the two sides FA, AD are equal to the two FB, BC, each to each ; and the base DF was proved equal to the base FC ; therefore the angle FAD is equal to the angle FBC: (i. 8.) again, it was proved that GA is equal to BH, and also AF to FB; therefore FA and AG are equal...
Side 94 - C, they are equiangular, and also have their sides about the equal angles proportionals (def. 1. 6.). Again, because B is similar to C, they are equiangular, and have their sides about the equal angles proportionals (def. 1. 6.) : therefore the figures A, B, are each of them equiangular to C, and have the sides about the equal angles of each of them, and of C, proportionals. Wherefore the rectilineal figures A and B are equiangular (1. Ax. 1.), and have their sides about the equal angles proportionals...