Geometry: A Formal Introduction

A Formal Introduction

The simplest polygon is a triangle. Given three noncollinear points A, B, and C, triangle ABC is the polygon formed by the sides ¯AB, ¯AC, and ¯BC. Figure 11.1 shows a picture of triangle ABC. The points A, B, and C will be referred to as the vertices of the triangle and we will refer to the triangle as ΔABC. The little triangle before the ABC is there to help you visualize it. A triangle has three sides (¯AB, ¯AC, and ¯BC) and three angles (∠ABC, ∠ACB, and ∠CAB). These angles are called the interior angles of ΔABC. Triangles can be classified by either the lengths of their sides or the measures of their interior angles.

Figure 11.1ΔABC.

Classifying Triangles by Their Angles or Their Sides

First let's classify triangles according to their size. Remember that two line segments are congruent if they have the same length. If all three sides of a triangle are congruent it is called an equilateral triangle. If only two of the sides are congruent, it is called an isosceles triangle. If no two sides are congruent, it is a scalene triangle. Triangles of each of these types are shown in Figure 11.2.

Figure 11.2Triangles can be classified by the number of congruent sides.

Triangles are also classified according to their angles. Recall that the four angle types are acute, right, obtuse, and straight. There is no such thing as a straight triangle, for reasons you will see soon. If a triangle has one obtuse angle, it is an obtuse triangle. A triangle cannot have more than one obtuse angle, because the sum of the measures of two obtuse angles is greater than the 180º allocated to a triangle. If a triangle has a right angle, it is a right triangle. A triangle cannot have more than one right angle, either. But a triangle can have more than one acute angle. In fact, every triangle has at least two acute angles, and if all three angles of a triangle are acute you have an acute triangle. If all three angles of a triangle are congruent, it is an equiangular triangle. Triangles of each of these types are shown in Figure 11.3.

Figure 11.3Triangles can be classified by angle size.

In a right triangle, the side opposite the right angle is called the hypotenuse of the triangle, and the other two shorter sides are called the legs of the triangle. In the right triangle shown in Figure 11.3, ¯AC and ¯BC are legs, and ¯AB is the hypotenuse of the right triangle ΔABC.

In general, a triangle is categorized by both its sides and its angles. For example, a triangle with two congruent sides and a right angle is called an isosceles right triangle. A triangle with one right angle but with no two sides congruent is called a right scalene triangle.

Sums of Interior Angles Are Cooking at 180º

You learned how to compute the sum of the interior angles of a polygon using the equation S = (n − 2) × 180º. You used this equation to show that the measures of the interior angles of a triangle add up to 180º. You can use this fact to prove some angle relationships and calculate some angle measures. If you (or anyone else) were to draw a triangle, measure its three angles, and add them together, the sum will always be 180º. It wouldn't matter what kind of triangle you drew, or how big it was. You would always get the same result when you added all three angles together: 180º. I will state this as a theorem, and use it freely from now on.

• Theorem 11.1: In a triangle, the sum of the measures of the interior angles is 180º

As a result of this theorem, you can establish two other results.

Suppose ΔABC is an equiangular triangle. What can you say about its angles? If all three angles of the triangle are congruent, then all three angles have the same measure. You don't know what the measure is, so you'll call it x. Because the sum of the measures of the interior angles of a triangle is 180º, you know that

• x + x + x = 180º
• 3x = 180º
• x = 60º

So, if you have an equiangular triangle, each interior angle measures 60º.

• Theorem 11.2: Each angle of an equiangular triangle measures 60º.

Here are examples of some algebraic problems that can be solved in this geometric environment.

• Example 1: Given that ΔNMQ is a right triangle, and ∠M is a right angle (see Figure 11.4), find m∠Q if m∠N = 44º.

Figure 11.4ΔNMQ is a right triangle, and ∠M is a right angle.

• Solution: The sum of the interior angles of ΔNMQ is 180º. Thus
• m∠M + m∠N + m∠Q = 180º.
• Because ∠M is a right angle, you know that m∠M = 90º. You were given that m∠N = 44º. Substituting into the above equation, you have:
• 90º + 44º + m∠Q = 180º
• 134º + m∠Q = 180º
• m∠Q = 46º

So m∠Q = 46º. Notice that the measures of ∠N and ∠Q add up to 90º. That means that ∠N and ∠Q are complementary. This is an interesting observation that you can prove for all right triangles.

Exterior Angle Relationships

When the sides of a triangle are extended, new angles appear on the scene. The angle formed by a side and an extension of the adjacent side is called an exterior angle of the triangle. In Figure 11.5, ∠ACD is an exterior angle of ΔABC. The two interior angles, ∠A and ∠B, can be described as nonadjacent interior angles of ∠ACD.

Figure 11.5∠ACD is an exterior angle of ΔABC.

Each vertex of a triangle has two exterior angles, for a grand total of six exterior angles. Figure 11.6 shows all six exterior angles of a triangle.

Figure 11.6Triangles have six exterior angles.

With these new angles, you can work out even more algebraic problems.

• Example 2: Suppose ΔMNQ has exterior angle ∠NQR , as in Figure 11.7. If m∠NQR = 117º and m∠M = 50º, find the measures of the other two interior angles of ΔMNQ.

Figure 11.7ΔMNQ with exterior angle ∠NQR.

• Solution: There are several angle pieces that fit together in this problem. ∠MQR is straight, so
• m∠MQR = 180º.
• Using the Angle Addition Postulate, you know that
• m∠MQN + m∠NQR = m∠MQR.
• Substituting in for ∠MQR and ∠NQR, you have
• m∠MQN + 117º = 180º
• m∠MQN = 63º

That's one angle down, and one to go. You know the measures of two of the three interior angles, so you can use Theorem 11.1 to find the measure of the third:

• m∠M + m∠N + m∠NQM = 180º
• m∠M + m∠N + m∠NQM = 180º
• 50º + m∠N + 63º = 180º
• m∠N = 67º

You have found the measure of the other two interior angles: m∠N = 67º and m∠MQN = 63º.

Notice that m∠M + m∠N = 50º + 67º = 117º. That's exactly the size of the exterior angle ∠NQR. This is not a coincidence. I'll state it as a theorem and let you prove it when you have finished reading this section.

• Theorem 11.3: The measure of an exterior angle of a triangle equals the sum of the measures of the two nonadjacent interior angles.