The Elements of Euclid, with Many Additional Propositions and Explanatory Notes. ... Part II.1876 |
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Page 2
... join CA , DEMONSTRATION . Because C is the cen- ter of the circle AEF , CA is equal to CE ( b ) ; but D is equal to CE ( c ) , therefore D is equal to CA ( α ) . SCHOLIUM . It should be observed that in the enunciation of the above ...
... join CA , DEMONSTRATION . Because C is the cen- ter of the circle AEF , CA is equal to CE ( b ) ; but D is equal to CE ( c ) , therefore D is equal to CA ( α ) . SCHOLIUM . It should be observed that in the enunciation of the above ...
Page 3
... join B and C. DEMONSTRATION . Because HAG touches the circle ABC , and AC is drawn from the point of contact , the angle HAC is equal to the angle B in the alternate segment of the circle ( d ) ; but the angle HAC is equal to the angle ...
... join B and C. DEMONSTRATION . Because HAG touches the circle ABC , and AC is drawn from the point of contact , the angle HAC is equal to the angle B in the alternate segment of the circle ( d ) ; but the angle HAC is equal to the angle ...
Page 4
... Join A and B , then because KAM and KBM are right angles ( d ) , the angles BAM and ABM are less than two right angles , A TI L B ( b ) I. 23 . c ) III . 17 . ( d ) III . 18 . G H E F Theor . attached to I. 29 . ( g ) I. 13 . I. 32 B ...
... Join A and B , then because KAM and KBM are right angles ( d ) , the angles BAM and ABM are less than two right angles , A TI L B ( b ) I. 23 . c ) III . 17 . ( d ) III . 18 . G H E F Theor . attached to I. 29 . ( g ) I. 13 . I. 32 B ...
Page 7
... Join BF , and from F draw GF and HF respectively B perpendicular to BC and CD ( c ) . H ( b ) Theor . attached to I. 29 . DEMONSTRATION . [ 1. ] In the triangle FCD , the angles FCD and FDC are equal , being the halves of equal angles ...
... Join BF , and from F draw GF and HF respectively B perpendicular to BC and CD ( c ) . H ( b ) Theor . attached to I. 29 . DEMONSTRATION . [ 1. ] In the triangle FCD , the angles FCD and FDC are equal , being the halves of equal angles ...
Page 8
... join AB , BC , CD , and DA , then ABCD is the square required . Because in the triangles B E D C ( a ) I. 4 . III . 31 . DEMONSTRATION . BEA and AED , BE and ED are equal , AE common to both , and at right angles to BD , the base AB is ...
... join AB , BC , CD , and DA , then ABCD is the square required . Because in the triangles B E D C ( a ) I. 4 . III . 31 . DEMONSTRATION . BEA and AED , BE and ED are equal , AE common to both , and at right angles to BD , the base AB is ...
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The Elements of Euclid, with Many Additional Propositions, and Explanatory ... Euclides No preview available - 2016 |
Common terms and phrases
altitude angle ABC angle BAC base ABC base DEF base EH bisected circle ABCD circle EFGH circumference common section cone Construction contained COROLLARY cylinder DEMONSTRATION duplicate ratio equal and similar equal angles equi equiangular equimultiples Euclid ex æquali fore four magnitudes fourth given circle given straight line greater ratio homologous homologous sides Hypoth inscribed join less LUDGATE HILL multiple opposite planes paral parallel parallelogram perpendicular polygon polyhedron prism PROPOSITION pyramid ABCG pyramid DEFH RALPH TATE reciprocally proportional rectangle rectilineal figure remaining angle right angles SCHOLIUM segments solid angle solid CD solid parallelopipeds sphere square on BD THEOREM THEOREM.-If third three plane angles three straight lines tiple triangle ABC triplicate ratio vertex vertex the point wherefore Woodcuts
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Page 10 - A KEY AND COMPANION to the above Book, forming an extensive repository of Solved Examples and Problems in Illustration of the various Expedients necessary in Algebraical Operations.
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