Elements of Geometry |
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Page 35
... difference BAD will have for its measure the half of BE minus the half of ED , or the half of BD . Therefore every inscribed angle has for its measure the half of he arc comprehended between its sides . 127. Corollary 1. All the angles ...
... difference BAD will have for its measure the half of BE minus the half of ED , or the half of BD . Therefore every inscribed angle has for its measure the half of he arc comprehended between its sides . 127. Corollary 1. All the angles ...
Page 51
... difference AC - AB — AE — AF , which gives BC = EF ; but , on account of the parallels , IG = BC , and DG EF , therefore HIGD is equal to the square described upon BC . These two parts being taken from the whole square , there remain ...
... difference AC - AB — AE — AF , which gives BC = EF ; but , on account of the parallels , IG = BC , and DG EF , therefore HIGD is equal to the square described upon BC . These two parts being taken from the whole square , there remain ...
Page 52
... difference of these lines ; therefore the rectangle AKLE = ( AB + BC ) × ( AB — BC ) . But this same rectangle is composed of two parts ABHE + BHLK , and the part BHLK is equal to the rectangle EDGF , for BH = DE , and BK = EF ...
... difference of these lines ; therefore the rectangle AKLE = ( AB + BC ) × ( AB — BC ) . But this same rectangle is composed of two parts ABHE + BHLK , and the part BHLK is equal to the rectangle EDGF , for BH = DE , and BK = EF ...
Page 54
... difference will be equal to double the rectangle BC x CD , or " 2 AB = AC + BC -2BC × CD . Demonstration . The proposition admits of two cases . 1. If the perpendicular fall within the triangle ABC , we shall have BD = BC — CD ; and ...
... difference will be equal to double the rectangle BC x CD , or " 2 AB = AC + BC -2BC × CD . Demonstration . The proposition admits of two cases . 1. If the perpendicular fall within the triangle ABC , we shall have BD = BC — CD ; and ...
Page 55
... difference will be equal to double the rectangle BC × CD , or , AB = AC + BC + 2BC × CD . Demonstration . The perpendicular cannot fall within the tri- angle ; for , if it should fall , for example , upon E , the triangle ACE would have ...
... difference will be equal to double the rectangle BC × CD , or , AB = AC + BC + 2BC × CD . Demonstration . The perpendicular cannot fall within the tri- angle ; for , if it should fall , for example , upon E , the triangle ACE would have ...
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Common terms and phrases
ABC fig adjacent angles altitude angle ACB angle BAC base ABCD bisect centre chord circ circular sector circumference circumscribed common cone consequently construction convex surface Corollary cube cylinder Demonstration diagonals diameter draw drawn equal angles equiangular equilateral equivalent faces figure formed four right angles frustum GEOM given point gles greater hence homologous sides hypothenuse inclination intersection isosceles triangle JOHN CRERAR LIBRARY join less Let ABC let fall line AC mean proportional measure the half meet multiplied number of sides oblique lines opposite parallelogram parallelopiped perimeter perpendicular plane MN polyedron prism produced proposition radii radius ratio rectangle regular polygon right angles Scholium sector segment semicircle semicircumference side BC similar solid angle sphere spherical polygons spherical triangle square described straight line tangent THEOREM three angles triangle ABC triangular prism triangular pyramids vertex vertices whence