 | Adrien Marie Legendre - 1819 - 574 σελίδες
...then cannot be supposed unequal to AC; therefore the sides AB, AC, opposite to the equal angles It. C, are equal. 484. Scholium. It is evident, from the...triangle to the middle of the base, is perpendicular lo this base, and divides the angle opposite into two equal parts. THEOREM. Fig. 332. 485. In any spherical... | |
 | Adrien Marie Legendre - 1822 - 394 σελίδες
...demonstration proves the angle BAD=DAC, and the angle BDA=ADC. Hence the two 172 last are right angles ; hence the arc drawn from the vertex of an isosceles spherical triangle to the middle of the base, is at right angles to that base, and bisects the opposite angle. PROPOSITION XVI. THEOREM. In a spherical... | |
 | Adrien Marie Legendre, John Farrar - 1825 - 294 σελίδες
...BC, are equal to the two AC, BC ; and the angle OBC contained by the first is equal to the angle ACB contained by the second. Consequently the two triangles...and divides the angle opposite into two equal parts. I s~ THEOREM. I ' 485. In any spherical triangle ABC (fig. 232), if the angle A is greater than the... | |
 | Adrien Marie Legendre, John Farrar - 1825 - 280 σελίδες
...cannot be supposed unequal to AC; therefore the sides AB, AC, opposite to the equal angles B, C, arc equal. 484. Scholium. It is evident, from the same...from the vertex of an isosceles spherical triangle to ihe middle of the base, is perpendicular to this base, and divides the angle opposite into two equal... | |
 | Adrien Marie Legendre - 1825 - 570 σελίδες
...DAC, and the angle BDA = ADC. Consequently the two last arc right angles ; therefore, the arc draxn from the vertex of an isosceles spherical triangle to the middle of iht base, is perpendicular to this base, and divides the angle opposite into two equal parts. THEOREM.... | |
 | Dionysius Lardner - 1828 - 434 σελίδες
...sides, they will be symmetrically equal, and the proposition has been already proved. (152.) Cor. Hence the arc drawn from the vertex of an isosceles spherical triangle to the point of bisection of the base, bisects the vertical angle, and is perpendicular to the base. * In... | |
 | Adrien Marie Legendre - 1836 - 394 σελίδες
...demonstration proves the angle BAD = DAC, and the angle BDA— ADC. Hence the two last are right angles ; hence the arc drawn from the vertex of an isosceles spherical triangle to the middle of the base, is at right angles to that base, and bisects the vertical angle. PROPOSITION XIV. THEOREM. In any spherical... | |
 | Benjamin Peirce - 1837 - 216 σελίδες
...Corollary. Also the angle ADB = ADC, and, therefore, each is a right angle ; and also DAB = DAC, that . is> The arc, drawn from the vertex of an isosceles spherical...triangle to the middle of the base, is perpendicular to the base, and bisects the angle at the vertex. 454. Corollary. An equilateral spherical triangle is... | |
 | Nathan Scholfield - 1845 - 542 σελίδες
...demonstration proves the angle BAD =DAC, and the angle BDA— ADC. Hence the two last are right angles ; hence the arc drawn from the vertex of an isosceles spherical triangle to the middle of the base, is at right angles to the base, and bisects the vertical angle. PROPOSITION XVI. THEOREM. In any spherical... | |
 | Nathan Scholfield - 1845 - 244 σελίδες
...demonstration proves the angle BAD =DAC, and the angle BDA=ADC. Hence the two last are right angles ; hence the arc drawn from the vertex of an isosceles spherical triangle to the middle of the base, is at right angles to the base, and bisects the vertical angle. PROPOSITION XVI. THEOREM. In any spherical... | |
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In any spherical triangle, the greater side is opposite the greater angle ; and conversely, the greater angle is opposite the greater side. 
