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.