Proposition XIX. Theorem.
|(97)||If in any triangle (B A C) one angle (B) be greater than another (C), the side (A C) which is opposite the greater angle is greater than the side A B, which is opposite to the less.|
For the side A C is either equal, or less, or greater than A B. It is not equal to A B, because the angle B would then be equal to C (V), which is contrary to the hypothesis.
It is not less than A B, because the angle B would then be less than C (XVIII), which is also contrary to the hypothesis.
Since therefore the side A C is neither equal to not less than A B, it is greater than it.
This proposition holds the same relation to the sixth, as the preceding does to the fifth. The four might be thus combined: one angle of a triangle is greater or less than another, or equal to it, according as the side opposed to the one is greater or less than, or equal to the side opposed to the other, and vice versa.
The student generally feels it difficult to remember which of the two, the eighteenth or nineteenth, is proved by construction, and which indirectly. By referring them to the fifth and sixth the difficulty will be removed.
Book I: Euclid, Book I (ed. Dionysius Lardner, 11th Edition, 1855)
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