II. ON TRIANGLES 25
triangles are congruent and, in particular, that the remaining elements are equal
as well.
III. Two congruent triangles (or, more generally, polygons) may differ in their
sense of rotation (20). In this case, they can only be superimposed by a motion
outside the plane. On the other hand, if the sense of rotation is the same, the two
polygons can be superimposed by moving them within the plane, as we will have
occasion to see later on.
25. The angle formed by a side of a convex polygon and the extension of the
next side is called an exterior angle of the polygon.
Theorem. An exterior angle of a triangle is greater than either of the non-
adjacent interior angles.
Figure 24
Let the triangle be ABC, and construct the exterior angle B AC (Fig. 24). We
claim that this angle is greater, for example, than the interior angle C. To show
this, we construct median BD, which we extend by its own length to a point E.
The point E being
inside2
angle B AC, this last must be greater than angle EAC.
But this last angle is precisely equal to C; indeed, triangles DAE, DBC are
congruent, having an equal angle between equal sides: the angles at D are equal
because they are vertical angles, and AD = DC, BD = DE by construction. Thus
exterior angle B AC is greater than interior angle C. QED
Exterior angle B AC is the supplement of interior angle A. Since angle C is
smaller than the supplement of A, the sum A + C is less than two right angles.
Our theorem can thus be restated: The sum of any two angles of a triangle is less
than two right angles. In particular, a triangle cannot have more than one right or
obtuse angle.
Theorem. In any triangle, the greater angle lies opposite the greater side.
In triangle ABC, suppose AB AC. we will show that C B. To show
this, we take, on AB, a length AD = AC (Fig. 25). It follows from the hypothesis
that point D is between A and B, and therefore angle ACD is less than C. But in
2Point
E is on the same side of line BAB as D (otherwise line DE, which has the common
point B with BAB , would have to cut it again between D and E, and this is impossible), and
thus on the same side as C. But B and E are on different sides of AC, which BE crosses at D.
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