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. Elements of Geometry - Page 4by Adrien Marie Legendre - 1825 - 224 pagesFull view - About this book
| Thomas Keith - Navigation - 1810 - 420 pages
...II. Hence 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 the angle FBC is equal... | |
| Adrien Marie Legendre - Geometry - 1819 - 208 pages
...equiangular, that is, it has its angles equal. 47. Scholium. From the equality of the triangles ABD, ACD, **it follows, that the angle BAD = DAC, and that 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, any one... | |
| Peter Nicholson - Architecture - 1823 - 596 pages
...C. 61. 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 - 224 pages
...the angle BAD = DAC, and that the angle BDA = ADC; therefore these two last are right angles, /fence **a straight line drawn from the vertex of an isosceles...angle. In an isosceles triangle, the base is that** sidewhich is not equal to one of the others. THEOREM. / W; 48. Reciprocally, if two angles of a triangle... | |
| Thomas Keith - Navigation - 1826 - 442 pages
...II. Hence 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 the angle FBC is... | |
| James Hayward - Geometry - 1829 - 172 pages
...PB, PC, AB, 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 line... | |
| Adrien Marie Legendre - Geometry - 1830 - 316 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 - 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... | |
| Thomas Keith - 1839
...Hence 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 the angle FBC is equal,... | |
| Adrien Marie Legendre - Geometry - 1841 - 235 pages
...From the equality of the triangles ABD, A CD. it follows, that the angle BAD = DAC, and that the anglf **BDA = ADC ; therefore these two last are right angles....triangle that is not isosceles, any one of its sides may** bt taken indifferently for a base ; and then its vertex is that of the opposite angle. In an isosceles... | |
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