| Автор | Russell, A. D. |
| Дата выпуска | 1945 |
| dc.description | In the figure, ABC is a triangle with B > C; AF is made equal to AC, AE bisects the angle A, and BDEG is a rectangle. It is easily seen that the angles marked α are equal, that B − α = C + α, and hence that α = ½ (B − C). Then: |
| Формат | application.pdf |
| Издатель | Cambridge University Press |
| Копирайт | Copyright © Edinburgh Mathematical Society 1945 |
| Название | A geometrical proof of |
| Тип | research-article |
| DOI | 10.1017/S0950184300000185 |
| Electronic ISSN | 0950-1843 |
| Print ISSN | 0950-1843 |
| Журнал | Edinburgh Mathematical Notes |
| Том | 35 |
| Первая страница | 9 |
| Последняя страница | 9 |
| Аффилиация | Russell A. D.; 12 Heugh Street, Falkirk. |
