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Constructions connected with Euclid VI., 3 and A, and the Circle of Apollonius

Published online by Cambridge University Press:  20 January 2009

R. F. Muirhead
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Let ABC be a triangle and AD the bisector of the angle A meeting BC in D. (Fig. 8.)

For simplicity let us suppose AB > AC.

From AB cut off AE = AC, and join DE. Then by Euclid I. 4 it follows that the triangles AED, ACD are equal.

This proof of VI. 3 has the merit of depending only on I. 4 and VI. 1. A similar proof for VI. A can be got by cutting off AE′ = AC from BA produced.

Type
Research Article
Information
Proceedings of the Edinburgh Mathematical Society , Volume 20 , February 1901 , pp. 67 - 69
Copyright
Copyright © Edinburgh Mathematical Society 1901