 | George Washington Hull - Geometry - 1807 - 408 pages
...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.... | |
 | Thomas Keith - Navigation - 1810 - 478 pages
...likewise equiangular, and the contrary. (I) COROLLARY III. A line drawn from the vertex of an isosceles triangle to the middle of the base, is perpendicular to, the base. For the two sides FB and BC are equal to the two sides FA and AC, and the angle FBC is equal to the... | |
 | Adrien Marie Legendre - Geometry - 1819 - 574 pages
...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. 485. In any spherical... | |
 | Adrien Marie Legendre - Geometry - 1822 - 394 pages
...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 - Geometry - 1825 - 294 pages
...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 ' 485. In any spherical... | |
 | Adrien Marie Legendre - 1825 - 570 pages
...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.... | |
 | Thomas Keith - Navigation - 1826 - 504 pages
...likewise equiangular, and the contrary. (I) COROLLARY III. A line drawn from the vertex of an isosceles triangle to the middle of the base, is perpendicular to the base. For the two sides FB and вс are equal to the two sides FA and AC, and the angle FBC is equal to the... | |
 | James Hayward - Geometry - 1829 - 228 pages
...angles opposite to the equal sides, are equal. 2. A straight line drawn from the summit of the isosceles triangle to the middle of the base, is perpendicular to the base, and bisects the angle ivhose vertex is at the summit. Fi%. 27. 46. If the triangle had been equilateral (fig. 27),... | |
 | Adrien Marie Legendre - Geometry - 1836 - 394 pages
...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 - Geometry - 1837 - 216 pages
...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...base, is perpendicular to the base, and bisects the angle at the vertex. 454. Corollary. An equilateral spherical triangle is also equiangular. 455. Theorem.... | |
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ADC ; the last two are therefore right angles ; hence 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 vertical angle. PROPOSITION XVI. 
