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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Resultat 1-5 av 70
Side 5
... third , EF , intersect them , it is pos- sible to describe two equal circles , each ful- filling the conditions of the problem , one on either side of the line EF ; 4 ° . If the three given lines intersect so as to form a triangle ...
... third , EF , intersect them , it is pos- sible to describe two equal circles , each ful- filling the conditions of the problem , one on either side of the line EF ; 4 ° . If the three given lines intersect so as to form a triangle ...
Side 10
... its base , A and ADC , are one - third of the vertical angle ACD . COROLLARY 3. PROBLEM . To divide a given right angle ( ABC ) into five equal parts . SOLUTION . In AB take any point A , and 10 ELEMENTS OF GEOMETRY .
... its base , A and ADC , are one - third of the vertical angle ACD . COROLLARY 3. PROBLEM . To divide a given right angle ( ABC ) into five equal parts . SOLUTION . In AB take any point A , and 10 ELEMENTS OF GEOMETRY .
Side 14
... third part of two right angles ( c ) ; and in the same manner it may be shown that the angle EGD is also the third part of two right angles . And because the straight line GC makes with EB the adjacent angles EGC , CGB , equal to two ...
... third part of two right angles ( c ) ; and in the same manner it may be shown that the angle EGD is also the third part of two right angles . And because the straight line GC makes with EB the adjacent angles EGC , CGB , equal to two ...
Side 15
... third part of the whole circumference , contains five of these parts ; in like manner the arc AB contains three of them , there- fore the arc BC contains two , and therefore the arc BE is the fifteenth part of the whole circumference ...
... third part of the whole circumference , contains five of these parts ; in like manner the arc AB contains three of them , there- fore the arc BC contains two , and therefore the arc BE is the fifteenth part of the whole circumference ...
Side 17
... third has to the fourth , when any equi- multiples whatsoever of the first and third are taken , and any equimultiples whatsoever of the second and fourth ; if the multi- ple of the first be less than that of the second , the multiple ...
... third has to the fourth , when any equi- multiples whatsoever of the first and third are taken , and any equimultiples whatsoever of the second and fourth ; if the multi- ple of the first be less than that of the second , the multiple ...
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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...