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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
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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.
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- Copyright © Edinburgh Mathematical Society 1901
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