A line drawn from the vertex of the right angle of a right triangle to the middle point of the hypotenuse divides the triangle into two isosceles triangles. Plane Geometry - Page 186by Edith Long, William Charles Brenke - 1916 - 276 pagesFull view - About this book
| William Frothingham Bradbury - Geometry - 1877 - 262 pages
...parallel sides ; and the part between the diagonals to half their difference. 155. The angle included by the perpendicular from the vertex of the right angle of a right triangle to the hypothenuse and the line from the same vertex bisecting the hypothenuse is equal to the difference... | |
| William Frothingham Bradbury - Geometry - 1880 - 260 pages
...parallel sides ; and the part between the diagonals to half their difference. 155i The angle included by the perpendicular from the vertex of the right angle of a right triangle to the hypothenuse and the line from the same vertex bisecting the hypothenuse is equal to the difference... | |
| George Albert Wentworth, George Anthony Hill - Geometry - 1894 - 150 pages
...vertex of the right angle. (Use the indirect method of proof with 191.) 4. Prove that the line joining the vertex of the right angle of a right triangle to the centre of the square constructed upon the hypotenuse bisects the right angle. 17. Propositions 190,... | |
| Andrew Wheeler Phillips, Irving Fisher - Geometry, Modern - 1896 - 276 pages
...contact with the third circle to their intersection which is nearest the large circle. 131. If CD is the perpendicular from the vertex of the right angle of a right triangle ABC, prove that the areas of the circles inscribed in the triangles ACD, BCD are proportional to the... | |
| George D. Pettee - Geometry, Plane - 1896 - 272 pages
...of AD, CD, BC, and AB, respectively. Prove that EFGH is a parallelogram. 90. If AD is drawn from A, the vertex of the right angle of a right triangle to the hypotenuse, so as to make the angle DA B = Z. B, it bisects the hypotenuse. 91. AB is the hypotenuse of a right... | |
| Andrew Wheeler Phillips, Irving Fisher - Geometry - 1896 - 554 pages
...contact with the third circle to their intersection which is nearest the large circle. 131. If CD is the perpendicular from the vertex of the right angle of a right triangle ABC, prove that the areas of the circles inscribed in the triangles ACD, BCD are proportional to the... | |
| Education - 1898 - 558 pages
...third side of the first is greater than the third side of the second. 3. Demonstrate: The line joining the vertex of the right angle of a right triangle to the center of the hypotenuse is equal to half the hypotenuse. NOTB. — In questions four and five construct... | |
| George Albert Wentworth - Geometry, Plane - 1899 - 276 pages
...perpendicular to the base. Z CBA =ZA, and Z CBD = Z D. (Why ?) .'. Z ABD = ZA + Z D. Ex. 44. A line drawn from the vertex of the right angle of a right triangle to the middle point of the hypotenuse divides the triangle into two isosceles triangles. Ex. 45. If the equal... | |
| George Albert Wentworth - Geometry - 1899 - 496 pages
...of the base is perpendicular to the base. ZCBA = ZA, and Z CBD = Z Z». (Why?) Ex. 44. A line drawn from the vertex of the right angle of a right triangle to the middle point of the hypotenuse divides the triangle into two isosceles triangles. Ex. 45. If the equal... | |
| Massachusetts. Board of Education - Education - 1899 - 782 pages
...by 10 feet and its breadth by 3 feet. Find the side of the square. GEOMETRY. Take 1. 1. Draw a line from the vertex of the right angle of a right triangle to the middle point of the hypothenuse. (a) What relation does this line bear to the halves of the hypothenuse... | |
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