 | Adrien Marie Legendre - 1819 - 574 sider
...angle 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, anil divides the vertical angle into two equal parts.... | |
 | Peter Nicholson - 1823 - 210 sider
...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.... | |
 | Adrien Marie Legendre, John Farrar - 1825 - 294 sider
...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 triangle, to the middle of the base, is perpendicular to that base, and divides the vertical angle into two equal parts.... | |
 | Adrien Marie Legendre, John Farrar - 1825 - 280 sider
...angle 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.... | |
 | James Hayward - 1829 - 228 sider
...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... | |
 | Adrien Marie Legendre - 1841 - 288 sider
...angle BAD = DAC, and 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.... | |
 | Euclides - 1845 - 546 sider
...intersection are together one-third of the squares of the sides of the triangle. 27. Prove that the square of any straight line drawn from the vertex of an isosceles triangle to the base, is less than the square of a side of the triangle by the rectangle contained by the segments of the base. 28. If from one of... | |
 | Peter Nicholson - 1856 - 518 sider
...COROLLARY 1. — 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.... | |
 | Robert Potts - 1860 - 380 sider
...the middle segment is equivalent to twice the rectangle under the extreme segments. V. 37. Prove that the square on any straight line drawn from the vertex...the rectangle contained by the segments of the base : and conversely. 38. If from one of the equal angles of an isosceles triangle a perpendicular be drawn... | |
 | Euclides - 1862 - 140 sider
...straight line which bisects the vertical angle of an isosceles triangle, bisects also the base. 3. The straight line drawn from the vertex of an isosceles triangle to the point of bisection of the base, bisects the vertical angle, and is perpendicular to the base. 4. In... | |
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Prove that the square on any straight line drawn from the vertex of an isosceles triangle to the base, is less than the square on a side of the triangle by the rectangle contained by the segments of the base : and conversely. 
