| Thomas Keith - Navigation - 1810 - 478 pages
...every equilateral triangle is 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... | |
| Adrien Marie Legendre - Geometry - 1819 - 574 pages
...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... | |
| Peter Nicholson - Architecture - 1823 - 210 pages
...— 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 a triangle... | |
| Adrien Marie Legendre, John Farrar - Geometry - 1825 - 294 pages
...and that the angle BDA = ADC; therefore these two last are right angles, /fence 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... | |
| Adrien Marie Legendre, John Farrar - Geometry - 1825 - 280 pages
...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... | |
| Thomas Keith - Navigation - 1826 - 504 pages
...every equilateral triangle is 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... | |
| James Hayward - Geometry - 1829 - 228 pages
...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 line... | |
| Thomas Keith - 1839 - 498 pages
...every equilateral triangle is likewise equiangular, and the contrary. (315) 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... | |
| Adrien Marie Legendre - Geometry - 1841 - 288 pages
...that the anglf 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... | |
| Peter Nicholson - Cabinetwork - 1856 - 482 pages
...Hence every equilateral triangle is also equiangular. в с 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 a triangle... | |
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