 | George Washington Hull - 1807 - 408 σελίδες
...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 σελίδες
...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.... | |
 | Adrien Marie Legendre - 1822 - 394 σελίδες
...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.... | |
 | Adrien Marie Legendre, John Farrar - 1825 - 294 σελίδες
...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... | |
 | Adrien Marie Legendre - 1836 - 394 σελίδες
...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.... | |
 | Benjamin Peirce - 1837 - 216 σελίδες
...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... | |
 | Adrien Marie Legendre - 1841 - 288 σελίδες
...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.... | |
 | Nathan Scholfield - 1845 - 542 σελίδες
...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.... | |
 | Nathan Scholfield - 1845 - 894 σελίδες
...proves the angle BAD =DAC, and the angle BDA=ADC. Hence the two last are ri^rht 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.... | |
 | Nathan Scholfield - 1845 - 244 σελίδες
...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.... | |
| |
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. 
