Elements of GeometryHilliard, Gray,, 1841 - 235 páginas |
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Página 7
... namely , the angle AD , the side AB = DE , and the side AC DF , we may thence infer , that the other three are also equal , namely , the angle B = E , the angle C = F , and the side BCEF . THEOREM . 38. Two triangles are equal , when a ...
... namely , the angle AD , the side AB = DE , and the side AC DF , we may thence infer , that the other three are also equal , namely , the angle B = E , the angle C = F , and the side BCEF . THEOREM . 38. Two triangles are equal , when a ...
Página 8
... namely , BC = EF , B = E , and C = F , we may thence infer , that the other three are also equal , namely , AB = DE , AC DF , and A = D. THEOREM . 40. One side of a triangle is less than the sum of the other two . Demonstration . The ...
... namely , BC = EF , B = E , and C = F , we may thence infer , that the other three are also equal , namely , AB = DE , AC DF , and A = D. THEOREM . 40. One side of a triangle is less than the sum of the other two . Demonstration . The ...
Página 9
... namely , A = D , BE , and CF. For , if the angle A were greater than the angle D , as the sides AB , AC , are equal to the sides DE , DF , each to each , the side BC would be greater than EF ( 42 ) ; and if the angle A were less than ...
... namely , A = D , BE , and CF. For , if the angle A were greater than the angle D , as the sides AB , AC , are equal to the sides DE , DF , each to each , the side BC would be greater than EF ( 42 ) ; and if the angle A were less than ...
Página 14
... namely , C'KIB CI , and KB ' = AKAI , since we have supposed AB ' = 2 AI = 2AK . Therefore the two triangles B'C'K , ACI , are equal ( 36 ) , and , consequently , the side C'B ' AC , and the angle B'C'KACB , and the angle KB'C ' — CAI ...
... namely , C'KIB CI , and KB ' = AKAI , since we have supposed AB ' = 2 AI = 2AK . Therefore the two triangles B'C'K , ACI , are equal ( 36 ) , and , consequently , the side C'B ' AC , and the angle B'C'KACB , and the angle KB'C ' — CAI ...
Página 22
... namely , AB CD , and AD CB , the equal sides will be parallel , and the figure will be a parallelogram . = = Demonstration . Draw the diagonal BD ; the two triangles ABD , BDC , have the three sides of the one equal to the three sides ...
... namely , AB CD , and AD CB , the equal sides will be parallel , and the figure will be a parallelogram . = = Demonstration . Draw the diagonal BD ; the two triangles ABD , BDC , have the three sides of the one equal to the three sides ...
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Términos y frases comunes
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 join less Let ABC let fall Let us suppose 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 third three angles triangle ABC triangular prism triangular pyramids vertex vertices whence
Pasajes populares
Página 67 - The square of the hypothenuse is equal to the sum of the squares of the other two sides ; as, 5033 402+302.
Página 9 - If two triangles have the three sides of the one equal to the three sides of the other, each to each, the triangles are congruent.
Página 65 - The areas of two triangles which have an angle of the one equal to an angle of the other are to each other as the products of the sides including the equal angles. A D A' Hyp. In triangles ABC and A'B'C', To prove AABC A A'B'C' A'B' x A'C ' Proof. Draw the altitudes BD and B'D'.
Página 160 - ABC (fig. 224) be any spherical triangle ; produce the sides AB, AC, till they meet again in D. The arcs ABD, ACD, will be...
Página 168 - In any spherical triangle, the greater side is opposite the greater angle ; and conversely, the greater angle is opposite the greater side.
Página 157 - CIRCLE is a plane figure bounded by a curved line, all the points of which are equally distant from a point within called the centre; as the figure ADB E.
Página 8 - Any side of a triangle is less than the sum of the other two sides...
Página 82 - The perimeters of two regular polygons of the same number of sides, are to each other as their homologous sides, and their areas are to each other as the squares of those sides (Prop.
Página 29 - Two equal chords are equally distant from the centre ; and of two unequal chords, the less is at the greater distance from the centre.
Página 182 - CD, &c., taken together, make up the perimeter of the prism's base : hence the sum of these rectangles, or the convex surface of the prism, is equal to the perimeter of its base multiplied by its altitude.