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 2 Point 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.

Purchased from American Mathematical Society for the exclusive use of nofirst nolast (email unknown) Copyright 2008 American Mathematical Society. Duplication prohibited. Please report unauthorized use to cust-serv@ams.org. Thank You! Your purchase supports the AMS' mission, programs, and services for the mathematical community.