| Adrien Marie Legendre - Geometry - 1819 - 574 pages
...DAC, and that the angle BDA — ADC ; therefore these two last are right angles. Hence, a straight line drawn from the vertex of an isosceles triangle, to the middle of the base, is perpendicular to that base, anil divides the vertical angle into two equal parts. In a triangle that is not isosceles,... | |
| Peter Nicholson - Architecture - 1823 - 210 pages
...COROLLARY 1. — Hence every equilateral triangle is also equiangular. 62. COROLLARY 2. — A straight line drawn from the vertex of an isosceles triangle to the middle of the base will bisect the vertical angle, and be perpendicular to the base. THEOREM 12. 63. If two angles of... | |
| Adrien Marie Legendre - Geometry - 1825 - 276 pages
...BAD = DAC, and that the angle BDA = ADC; therefore these two last are right angles. Hence a straight line drawn from the vertex of an isosceles triangle, to the middle of the base, is perpendicular to that base, and divides the vertical angle into two equal parts. . In a triangle that is not isosceles,... | |
| Adrien Marie Legendre, John Farrar - Geometry - 1825 - 280 pages
...— DAC, and that the angle BDA = ADC ; therefore these two last are right angles. Hence a straight line drawn from the vertex of an isosceles triangle, to the middle of the base, is perpendicular to that base, and divides the vertical angle into two equal parts. In a triangle that is not isosceles,... | |
| Thomas Keith - Navigation - 1826 - 504 pages
...the triangle ADC is equal to the triangle ВЕС (E. 142.); consequently AE+ EC is equal to ED + DC. viz. equal sides are opposite to equal angles. (H)...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 angle FAC, therefore the... | |
| James Hayward - Geometry - 1829 - 218 pages
...AC ; PB and PC will be equal (41), and AB and AC will therefore be equal ; and AG will be a straight line drawn from the vertex of an isosceles triangle to the middle of the base ; that is — BC is perpendicular to the oblique line AG, when it is perpendicular to the straight... | |
| Adrien Marie Legendre - Geometry - 1830 - 344 pages
...that the angle BAD is equal to DAC, and BDA to ADC ; hence the latter two are right angles ; hence the line drawn from the vertex of an isosceles triangle to the middle point of its base, is perpendicular to that base, and divides the angle at the vertex into two equal... | |
| Adrien Marie Legendre - Geometry - 1836 - 394 pages
...angle BAD, is equal to DAC, and BDA to ADC, hence the latter two are right angles ; therefore, the line drawn from the vertex of an isosceles triangle to the middle point of its base, is perpendicular to the base, and divides the angle at the vertex into two equal... | |
| Benjamin Peirce - Geometry - 1837 - 216 pages
...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 also equiangular.... | |
| Thomas Keith - 1839 - 498 pages
...(314) COROLLARY II. Hence every equilateral triangle is likewise equiangular, and the contrary. (315) COROLLARY III. A line drawn from the vertex of an...therefore the angle CFB is equal to the angle CFA (310) ; but CFA and CFB are equal to two right angles (294), therefore CF is perpendicular to AB. (316)... | |
| |