Elements of Geometry |
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Page 45
... parallelograms ABCD , ABEF ; since they are supposed to have the same altitude , the sides DC , FE , opposite to the bases , will be situated in a line parallel to AB ( 78 ) . Now , by the nature of a parallelogram , AD = BC ( 81 ) ...
... parallelograms ABCD , ABEF ; since they are supposed to have the same altitude , the sides DC , FE , opposite to the bases , will be situated in a line parallel to AB ( 78 ) . Now , by the nature of a parallelogram , AD = BC ( 81 ) ...
Page 46
... parallelogram ABEF ; and if , from the same quadrilateral ABED , the triangle CBE , equal to the former , be taken , there will remain the parallelogram ABCD ; therefore the two parallelograms ABCD , ABEF , which have the same base and ...
... parallelogram ABEF ; and if , from the same quadrilateral ABED , the triangle CBE , equal to the former , be taken , there will remain the parallelogram ABCD ; therefore the two parallelograms ABCD , ABEF , which have the same base and ...
Page 49
... parallelogram ABCD . 175. Corollary . Parallelograms of the same base are to each other as their altitudes , and parallelograms of the same altitude are to each other as their bases ; for , A , B , C , being any three magnitudes ...
... parallelogram ABCD . 175. Corollary . Parallelograms of the same base are to each other as their altitudes , and parallelograms of the same altitude are to each other as their bases ; for , A , B , C , being any three magnitudes ...
Page 50
... parallelogram ADKL , and has for its measure EF × AL . But AL = DK ; and , since the triangle IBL is equal to the triangle KCI , the side BL = CK ; therefore AB + CD = AL + DK = 2 AL ; thus AL is half the sum of the sides AB , CD ; and ...
... parallelogram ADKL , and has for its measure EF × AL . But AL = DK ; and , since the triangle IBL is equal to the triangle KCI , the side BL = CK ; therefore AB + CD = AL + DK = 2 AL ; thus AL is half the sum of the sides AB , CD ; and ...
Page 55
... parallelogram the sum of the squares of the sides is equal to the sum of the squares of the diagonals . For the diagonals AC , BD ( fig . 113 ) , mutually hisect each Fig . 113 . other in the point E ( 88 ) , and the triangle ABC gives ...
... parallelogram the sum of the squares of the sides is equal to the sum of the squares of the diagonals . For the diagonals AC , BD ( fig . 113 ) , mutually hisect each Fig . 113 . other in the point E ( 88 ) , and the triangle ABC gives ...
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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