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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Página 2
... diameter of the circle . SOLUTION . Draw a diameter BC of the circle ; and if this be equal to the given line D , the problem is solved ; but if it is not , take in it the segment CE equal to D ( a ) , and from C as a center , with the ...
... diameter of the circle . SOLUTION . Draw a diameter BC of the circle ; and if this be equal to the given line D , the problem is solved ; but if it is not , take in it the segment CE equal to D ( a ) , and from C as a center , with the ...
Página 2
... diameter of the circle . SOLUTION . Draw a diameter BC of the circle ; and if this be equal to the given line D , the problem is solved ; but if it is not , take in it the segment CE equal to D ( a ) , and from C as a center , with the ...
... diameter of the circle . SOLUTION . Draw a diameter BC of the circle ; and if this be equal to the given line D , the problem is solved ; but if it is not , take in it the segment CE equal to D ( a ) , and from C as a center , with the ...
Página 8
... diameters AC , BD , at right angles to each other ; and join AB , BC , CD , and DA , then ABCD is the square required ... diameter , ABD is a semicircle , and therefore the angle BAD is a right angle ( b ) , and the quadrilateral figure ...
... diameters AC , BD , at right angles to each other ; and join AB , BC , CD , and DA , then ABCD is the square required ... diameter , ABD is a semicircle , and therefore the angle BAD is a right angle ( b ) , and the quadrilateral figure ...
Página 14
... diameter AD . From D as a center with the radius DG describe a circle GEHC , join EG and GC , and produce them to the points B and F. Join AF , FE , ED , DC , CB , BA , with straight lines , and they will form an equilateral and equian ...
... diameter AD . From D as a center with the radius DG describe a circle GEHC , join EG and GC , and produce them to the points B and F. Join AF , FE , ED , DC , CB , BA , with straight lines , and they will form an equilateral and equian ...
Página 98
... diameter . B 江湖 K H F E DEMONSTRATION . For , if not , let the parallelogram BD have its dia- meter AHC in a different straight line from AF the diameter of the parallelogram EG , and let GF meet AHC in H ; through H draw HK parallel ...
... diameter . B 江湖 K H F E DEMONSTRATION . For , if not , let the parallelogram BD have its dia- meter AHC in a different straight line from AF the diameter of the parallelogram EG , and let GF meet AHC in H ; through H draw HK parallel ...
Otras ediciones - Ver todas
The Elements of Euclid: With Many Additional Propositions, & Explanatory ... Euclid Sin vista previa disponible - 2023 |
The Elements of Euclid: With Many Additional Propositions, and Explanatory ... Euclid Sin vista previa disponible - 2013 |
Términos y frases comunes
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
Pasajes populares
Página 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.
Página 75 - ... if the segments of the base have the same ratio which the other sides of the triangle have to one another...
Página 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.
Página 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.
Página 198 - Equiangular parallelograms have to one another the ratio which is compounded of the ratios of their sides.
Página 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.
Página 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.
Página 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.
Página 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...
Página 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...