A geometrical proof of
Russell, A. D.; Russell A. D.; 12 Heugh Street, Falkirk.
Журнал:
Edinburgh Mathematical Notes
Дата:
1945
Аннотация:
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:
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