 All the interior angles of any rectilineal figure, together with four right angles, are equal to twice as many right angles as the figure has sides.  The Elements of Euclid - Pagina 19door Euclid - 1838 - 416 pagina’sVolledige weergave - Over dit boek
 | Euclides - 1855 - 270 pagina’s
...triangles are equal to twice as many right angles as the figure has sides. Therefore all the angles of the figure together with four right angles are equal to...twice as many right angles as the figure has sides. СOR. 2. — All the exterior angles of any rectilineal figure, made by producing the sides successively... | |
 | William Mitchell Gillespie - 1855 - 436 pagina’s
...proposition of Geometry, that in any figure bounded by straight lines, the sum of all the interior angles is equal to twice as many right angles, as the figure has sides less two ; since the figure can be divided into that number of triangles. Hence this common rule. "... | |
 | Cambridge univ, exam. papers - 1856 - 200 pagina’s
...also be equal. Prove this also without construction, by superposition. 3. Prove that all the internal angles of any rectilineal figure, together with four...twice as many right angles as the figure has sides; and that all the external angles are together equal to four right angles. In what sense are these propositions... | |
 | Euclides - 1856 - 168 pagina’s
...vertex of the triangles ; that is, together with four right angles. Therefore all the angles of the figure, together with four right angles, are equal...twice as many right angles as the figure has sides. XVI. If two triangles have two sides of the one equal to two sides of the other, each to each, and... | |
 | 1856 - 428 pagina’s
...triangles thus formed are equal to all the angles of the figure (Const.) ; therefore all the angles of the figure, together with four right angles, are equal to twice as many right angles as the figure nas sides (Лх. 1). QED The demonstration of Euclid's Cor. II. viz. "that all the pxterior angles... | |
 | Henry James Castle - 1856 - 220 pagina’s
...that these angles are the exterior angles of an irregular polygon ; and as the sum of all the interior angles are equal to twice as many right angles, as the figure has sides, wanting four ; and as the sum of all the exterior, together with all the interior angles, are equal... | |
 | William Mitchell Gillespie - 1856 - 478 pagina’s
...proposition of Geometry, that in any figure bounded by straight lines, the sum of all the interior angles is equal to twice as many right angles, as the figure has sides less two ; since the figure can be divided into that number of triangles. Hence this common rule. "... | |
 | William Mitchell Gillespie - 1857 - 538 pagina’s
...proposition of Geometry, that in any figure bounded by straight lines, the sum of all the interior angles is equal to twice as many right angles, as the figure has sides less two ; since the figure can be divided into that number of triangles. Hence this common rule. "... | |
 | W J. Dickinson - 1879 - 44 pagina’s
...produced to meet, the angles formed by these lines, together with eight right angles, are together equal to twice as many right angles as the figure has sides. Same proposition. ABC is a triangle right-angled at A, and the angle B is double of the angle C. Show... | |
 | Isaac Sharpless - 1879 - 282 pagina’s
...But ACD+ACB = 2R; BAC+ABC+ACB = 2R. Corollary 1.—All the interior angles of a polygon are together equal to' twice as many right angles as the figure has sides, minus four right angles. Let ABODE be a polygon, and let n represent the number of its sides. Draw... | |
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