| Adrien Marie Legendre - 1819 - 208 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 - 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 - 1825 - 280 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 - 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 - 1826 - 442 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 - 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 - 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 - 1836 - 359 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 - 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)... | |
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