Elements of Geometry and Trigonometry: From the Works of A. M. Legendre |
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Page 14
... common point A , is called the ver- A- -B tex . An angle is designated by naming its sides , or some- times by simply naming its vertex ; thus , the above is called the angle BAC , or simply , the angle A. 11. When one straight line ...
... common point A , is called the ver- A- -B tex . An angle is designated by naming its sides , or some- times by simply naming its vertex ; thus , the above is called the angle BAC , or simply , the angle A. 11. When one straight line ...
Page 22
... common A angle . ACE ( A. 3 ) , there re- mains , ACD ECB . E B In like manner , we find , ACD + ACE = ACD + DCB ; and , taking away the common angle ACD ,. we have , ACE DCB . Hence , the proposition is proved . Cor . 1. If one of the ...
... common A angle . ACE ( A. 3 ) , there re- mains , ACD ECB . E B In like manner , we find , ACD + ACE = ACD + DCB ; and , taking away the common angle ACD ,. we have , ACE DCB . Hence , the proposition is proved . Cor . 1. If one of the ...
Page 23
... common , they will coincide throughout their whole extent , and form one and the same line . Let A and B be two points common to two lines : then will the lines coincide throughout . E A B C -D Between A and B they must coincide ( A. 11 ) ...
... common , they will coincide throughout their whole extent , and form one and the same line . Let A and B be two points common to two lines : then will the lines coincide throughout . E A B C -D Between A and B they must coincide ( A. 11 ) ...
Page 24
... common angle DCA , there re- mains , DCB = DCE , which is impossible , since a part cannot be equal to the whole ( A. 8 ) . Hence , CB must be the prolongation of AC ; which was to be proved . PROPOSITION V. THEOREM . If two triangles ...
... common angle DCA , there re- mains , DCB = DCE , which is impossible , since a part cannot be equal to the whole ( A. 8 ) . Hence , CB must be the prolongation of AC ; which was to be proved . PROPOSITION V. THEOREM . If two triangles ...
Page 29
... common part AB , there remains ( A. 5 ) , BC > EF . 2o . When G is on BC . In this case , it is obvious that GC is less than BC ; or , since GC EF , we have , D BC > EF . B E 3 ° . When G is within the triangle ABC . From Proposition ...
... common part AB , there remains ( A. 5 ) , BC > EF . 2o . When G is on BC . In this case , it is obvious that GC is less than BC ; or , since GC EF , we have , D BC > EF . B E 3 ° . When G is within the triangle ABC . From Proposition ...
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Common terms and phrases
AB² ABCD altitude apothem Applying logarithms centre chord circle circumference cone consequently convex surface cosec Cosine Cotang cylinder demonstrated in Book denote diameter distance divided draw edges Equation feet find the area Find the logarithmic following RULE frustum given angle greater hence homologous hypothenuse included angle inscribed intersection less Let ABC linear units log cot log sin lower base lune mantissa multiplied number of sides opposite parallel parallelogram parallelopipedon perpendicular plane MN polar triangle polyedral angle polyedron principle demonstrated prism proportional PROPOSITION proved pyramid quadrant radii radius rectangle regular polygon right angles right-angled triangle Scholium segment similar six right slant height solution sphere spherical angle spherical excess spherical polygon spherical triangle square straight line subtracting Tang tangent THEOREM triangle ABC triangular prism upper base vertex volume whence write the following