 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.  Plane and Solid Geometry - Side 385av George Albert Wentworth - 1904 - 473 siderUten tilgangsbegrensning - Om denne boken
 | George Washington Hull - 1807 - 408 sider
...ZXBCare mutually equilateral. Hence they are mutually equiangular. § 603 Therefore ZA = /. BQED 608. COR. — The arc of a great circle drawn from the vertex of an isosceles triangle to the middle of the base bisects the vertical angle and is perpendicular to the base. EXERCISES.... | |
 | Adrien Marie Legendre - 1819 - 574 sider
...BAD = DAC, and the angle BDA = ADC. Consequently the two last are right angles ; therefore, the arc drawn from the vertex of an isosceles spherical triangle to the middle of the base, is perpendicular lo this base, and divides the angle opposite into two equal parts. THEOREM. Fig. 332.... | |
 | Adrien Marie Legendre - 1822 - 394 sider
...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 sider
...BAD — DAC, and the angle BDA = ADC. Consequently the two last are right angles ; therefore, the arc drawn from the vertex of an isosceles spherical triangle to the middle of the base, is perpendicular to this base, and divides the angle opposite into two equal parts. I s~ THEOREM. I... | |
 | Adrien Marie Legendre - 1825 - 570 sider
...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.... | |
 | Adrien Marie Legendre - 1836 - 394 sider
...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 sider
...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... | |
 | Adrien Marie Legendre - 1841 - 288 sider
...BAD = DAC, and the angle BDA — ADC. Consequently the two last are right angles ; therefore, the arc drawn from the vertex of an isosceles spherical triangle to the middle of the base, is perpendicular to this base, and divides the angle opposite into two equal parts. THEOREM. . 232.... | |
 | Nathan Scholfield - 1845 - 244 sider
...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 - 542 sider
...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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