Lesson Explainer: Introduction to Inequalities in Triangles | Nagwa Lesson Explainer: Introduction to Inequalities in Triangles | Nagwa

Lesson Explainer: Introduction to Inequalities in Triangles Mathematics

In this explainer, we will learn how to define the concept of inequality in general and the axioms of inequality and compare between the measures of angles given in a triangle.

Before we begin discussing inequalities in triangles, we first need to recall what inequalities represent and how we can manipulate them. First, we recall that we use the symbol < to represent “less than” and > to represent “greater than.” For example, we can say that 5>3 since 5 is a larger value than 3.

We can use inequalities to represent a comparison between any two numbers. This is particularly useful if the numbers represent a quantity such as distance. To see why this is useful, consider the following diagram:

We can see that the line segment 𝐶𝐷 is longer than the line segment 𝐴𝐵, but writing this sentence out in full is not as succinct as writing the inequality 𝐶𝐷>𝐴𝐵.

This extends to any quantities represented by numbers. For example, consider the following two angles:

We can see in the diagram that the second angle is larger than the first angle. However, we need to be slightly careful. Remember that we describe the measure of the angle by a number and not the actual angle itself. So, when we are comparing the sizes of the angles, we should be comparing their measures. We can write this as 𝑚𝐸𝐷𝐹>𝑚𝐶𝐴𝐵.

We also recall that we can manipulate these inequalities to find equivalent inequalities. This can be useful in solving algebraic inequalities, such as finding all the values of 𝑥 that satisfy the inequality 𝑥+1<2. To do this, we subtract 1 from both sides of the inequality to get 𝑥<1. However, this can also be useful in comparing lengths, angle measures, or any other quantities.

For example, consider the following angles.

We see these two angles together make up less than a straight angle, and we know that the measure of a straight angle is 180. Therefore, we can conclude using the diagram that the sum of the measures of these angles is less than 180. We write this as 𝑚𝐴𝐵𝐶+𝑚𝐶𝐵𝐷<180.

To manipulate inequalities, we apply the same operation to both sides of the inequality using rules called the axioms of inequalities. We define the axioms of inequalities as follows:

Definition: Axioms of Inequalities

For any real numbers 𝑎, 𝑏, 𝑐, and 𝑑, we have the following:

  • We can compare the sizes of any two real numbers. In other words, either 𝑎>𝑏,𝑎<𝑏, or 𝑎=𝑏.
  • Inequalities and equalities are transitive relations on the set of real numbers. This means that if 𝑎>𝑏 and 𝑏>𝑐, then we must have that 𝑎>𝑐. This is called the transitivity axiom.
  • Adding a constant to or subtracting a constant from both sides of an inequality gives an equivalent inequality. For example, if 𝑎>𝑏, then 𝑎+𝑐>𝑏+𝑐 and 𝑎𝑐>𝑏𝑐.
  • Multiplying or dividing both sides of an inequality by a positive constant yields an equivalent inequality. For example, if 𝑎>𝑏 and 𝑐>0, then 𝑎𝑐>𝑏𝑐 and 𝑎𝑐>𝑏𝑐.
  • We can combine the ideas of the transitivity axiom with the other axioms. For example, if 𝑎>𝑏 and 𝑐>𝑑, then 𝑎+𝑐>𝑏+𝑑.

We can apply these axioms to any inequality to help solve inequalities or rewrite inequalities in a way that highlights certain relationships. For example, we can recall that the sum of the measures of the interior angles in a triangle is 180. If we consider a right triangle 𝐴𝐵𝐶, we can prove that the nonright angles must be acute.

First, we note that 90+𝑚𝐴𝐶𝐵+𝑚𝐴𝐵𝐶=180. We subtract 90 from both sides to get 𝑚𝐴𝐶𝐵+𝑚𝐴𝐵𝐶=18090𝑚𝐴𝐶𝐵+𝑚𝐴𝐵𝐶=90.

We know that we can always compare the sizes of real numbers, so we can compare 𝑚𝐴𝐶𝐵 to 90 and 𝑚𝐴𝐵𝐶 to 90. Since their measures add to 90 and they are both positive, neither angle can have a measure greater than or equal to 90. Hence, 𝑚𝐴𝐶𝐵<90,𝑚𝐴𝐵𝐶<90.

Let’s now see some examples of comparing geometric quantities using inequalities and using the axioms of inequalities to prove geometric relationships.

Example 1: Forming Inequalities in a Triangle

Look at the figure.

Use <, =, or > to fill in the blank in the following.

  • 𝑚𝐴𝐶𝐷𝑚𝐴𝐷𝐸.
  • 𝑚𝐴𝐶𝐷𝑚𝐴𝐵𝐶.
  • 𝑚𝐴𝐷𝐶𝑚𝐴𝐶𝐵.
  • 𝑚𝐴𝐷𝐸𝑚𝐶𝐴𝐷.

Answer

To compare the measures of these angles, we will start by finding the measures of the angles in the diagram. We can do this by recalling two facts:

  • The sum of the measures of the internal angles in a triangle is 180.
  • The sum of the measures of the angles that make up a straight angle is 180.

We can find 𝑚𝐴𝐶𝐵 by noting 𝐴𝐶𝐵 is an internal angle of triangle 𝐴𝐵𝐶. Thus, 𝑚𝐴𝐶𝐵+17+24=180𝑚𝐴𝐶𝐵=1801724=139.

Next, we can find 𝑚𝐴𝐶𝐷 by noting 𝐴𝐶𝐷 combines with 𝐴𝐶𝐵 to make a straight angle. Therefore, 𝑚𝐴𝐶𝐷+𝑚𝐴𝐶𝐵=180.

Substituting 𝑚𝐴𝐶𝐵=139 into the equation and solving yields 𝑚𝐴𝐶𝐷+139=180𝑚𝐴𝐶𝐷=180139=41.

We now have the measures of two of the internal angles in triangle 𝐴𝐶𝐷, so we can determine the measure of the final internal angle. We have 𝑚𝐴𝐷𝐶+71+41=180𝑚𝐴𝐷𝐶=1807141=68.

Finally, we see that 𝐴𝐷𝐸 combines with 𝐴𝐷𝐶 to make a straight angle, so their measures sum to 180. Thus, 𝑚𝐴𝐷𝐸+68=180𝑚𝐴𝐷𝐸=18068=112.

We can add these angle measures onto the diagram to get the following:

We can now compare the measures of all of these angles.

Firstly, 𝑚𝐴𝐶𝐷=41 and 𝑚𝐴𝐷𝐸=112, so 𝑚𝐴𝐶𝐷<𝑚𝐴𝐷𝐸.

Secondly, 𝑚𝐴𝐶𝐷=41 and 𝑚𝐴𝐵𝐶=17, so 𝑚𝐴𝐶𝐷>𝑚𝐴𝐵𝐶.

Thirdly, 𝑚𝐴𝐷𝐶=68 and 𝑚𝐴𝐶𝐵=139, so 𝑚𝐴𝐷𝐶<𝑚𝐴𝐶𝐵.

Finally, 𝑚𝐴𝐷𝐸=112 and 𝑚𝐶𝐴𝐷=71, so 𝑚𝐴𝐷𝐸>𝑚𝐶𝐴𝐷.

In the previous example, we compared the measures of the interior and exterior angles in a triangle by using their measures, the measure of a straight angle, and the sum of the measures of the interior angles in a triangle.

It is worth noting we can use this same reasoning to show a useful inequality that holds in general. Consider any triangle 𝐴𝐵𝐶 and any of its exterior angles, for example, the exterior angle at 𝐵 in the following diagram.

We know that the sum of the measures of the interior angles in a triangle is 180. So, 𝑎+𝑏+𝑐=180.

Similarly, we know that the exterior and interior angles at 𝐵 make up a straight angle, so the sum of their measures is 180. Thus, 𝑏+𝑚𝐶𝐵𝐷=180.

Since the left-hand sides of both equations are equal to 180, we can equate the left-hand sides of the equations to get 𝑎+𝑏+𝑐=𝑏+𝑚𝐶𝐵𝐷.

We can subtract 𝑏 from both sides of the equation to get 𝑎+𝑐=𝑚𝐶𝐵𝐷.

We now note that 𝑎, 𝑐, 𝑚𝐶𝐵𝐷>0 since they are measures of angles that are not the zero angle. Since 𝑎 and 𝑐 are positive, we can subtract either positive number from the left-hand side of the equation to make it smaller. Hence, 𝑎<𝑚𝐶𝐵𝐷𝑐<𝑚𝐶𝐵𝐷.and

We have proven that the measure of an exterior angle in any triangle is greater than the measure of either of the two nonadjacent interior angles of the triangle. We state this result formally as follows.

Property: Comparing the Measures of the Interior and Exterior Angles in a Triangle

The measure of any exterior angle in a triangle 𝐴𝐵𝐶 is greater than the measure of either nonadjacent interior angle in the triangle.

In our next example, we will use this property to compare all of the angles in a diagram with measures smaller than a given angle.

Example 2: Identifying Angles in a Triangle Less Than Another Angle

In the given figure, which angles must have smaller measures than 𝑚1?

Answer

In this question, we need to identify all of the angles whose measures are less than the measure of 1. However, we are not given the measures of any of these angles, so instead, we will need to use the diagram and the relationships between the angles to compare their measures.

In particular, we recall that the measure of any exterior angle in a triangle 𝐴𝐵𝐶 is greater than the measure of either nonadjacent interior angle in the triangle.

We see that 1 is an exterior angle in the following triangle.

Hence, its measure is greater than those of the two nonadjacent interior angles in this triangle, which are 4 and 5.

We can show that none of the other angles have to have angle measures smaller than 𝑚1 by considering an example. Let the triangle in red be an isosceles triangle and the measure of angle 1 be 60, so that we have the following:

We can add any line we want to construct the 4 missing angles. Since we have an isosceles triangle, we recall that the angle bisector and perpendicular bisector of the base are the same line. Adding this line in gives us the following:

We see that 𝑚6=𝑚7=90, which is larger than the measure of 1. So, these do not have smaller measures than 1. Next, 𝑚2=𝑚3, and we see that 1, 2, and 3 combine to make a straight angle. Thus, their measures add to 180: 𝑚1+𝑚2+𝑚3=180.

Substituting 𝑚1=60 and 𝑚2=𝑚3 yields 60+𝑚2+𝑚2=180.

Subtracting 60 from both sides of the equation and simplifying then gives 2𝑚2=18060𝑚2=60.

Therefore, 𝑚2=𝑚3=60.

This is equal in measure to 1. So, these do not have smaller measures than 1.

Whilst the red triangle is not necessarily isosceles, we have demonstrated that there are circumstances in which the remaining angles may not have a measure smaller than 𝑚1.

Hence, only 4 and 5 must have measures smaller than 𝑚1.

In our next example, we will order the measures of the interior angles in a triangle using the measures of an interior and exterior angle.

Example 3: Using Inequalities to Order the Angles in a Triangle

Which of the following inequalities is correct?

  1. 𝑚𝐴𝐵𝐶<𝑚𝐵𝐴𝐶<𝑚𝐴𝐶𝐵
  2. 𝑚𝐴𝐵𝐶>𝑚𝐵𝐴𝐶>𝑚𝐴𝐶𝐵
  3. 𝑚𝐵𝐴𝐶>𝑚𝐴𝐵𝐶>𝑚𝐴𝐶𝐵
  4. 𝑚𝐵𝐴𝐶<𝑚𝐴𝐵𝐶<𝑚𝐴𝐶𝐵
  5. 𝑚𝐴𝐶𝐵<𝑚𝐴𝐵𝐶<𝑚𝐵𝐴𝐶

Answer

We can answer this question by determining the measures of each of the angles in the triangle.

First, we recall that the sum of the measures of angles that make up a straight angle is 180. So, 109+𝑚𝐵𝐴𝐶=180.

Subtracting 109 from both sides of the equation yields 𝑚𝐵𝐴𝐶=180109=71.

We then recall that the sum of the measures of the interior angles in a triangle is 180. Thus, 71+𝑚𝐴𝐵𝐶+29=180𝑚𝐴𝐵𝐶=1807129=80.

Therefore, we can see that 𝑚𝐴𝐵𝐶>𝑚𝐵𝐴𝐶 and 𝑚𝐵𝐴𝐶>𝑚𝐴𝐶𝐵. We can combine these into one inequality written as 𝑚𝐴𝐵𝐶>𝑚𝐵𝐴𝐶>𝑚𝐴𝐶𝐵, which is option B.

In our next example, we will use the properties of isosceles triangles to compare the measures of two angles in a given diagram.

Example 4: Proving an Inequality in a Triangle

Use <, =, or > to complete the statement: If 𝑚𝐴𝐶𝐵=62 and 𝑚𝐴=57, then 𝑚𝐴𝐵𝐷𝑚𝐴𝐶𝐷.

Answer

We start by noticing that 𝐶𝐷=𝐵𝐷, so triangle 𝐵𝐶𝐷 is an isosceles triangle. We then recall that the angles opposite the congruent sides in an isosceles triangle are congruent, so 𝑚𝐵𝐶𝐷=𝑚𝐷𝐵𝐶. We can add this congruence and the given angle measures to the diagram as follows:

We can determine the measure of the angle at 𝐵 by recalling that the sum of the measures of the interior angles in a triangle is 180. So, 𝑚𝐴+𝑚𝐴𝐵𝐶+𝑚𝐴𝐶𝐵=18057+𝑚𝐴𝐵𝐶+62=180𝑚𝐴𝐵𝐶=61.

We can then add this onto the diagram as shown.

If we call the measure of the congruent angles 𝑥, then by comparing the angles at 𝐵 and 𝐶 with the smaller angles at these vertices we can see that 𝑥+𝑚𝐴𝐶𝐷=62,𝑥+𝑚𝐴𝐵𝐷=61.

Adding 1 to both sides of the second equation gives 𝑥+𝑚𝐴𝐵𝐷+1=62.

Now, the left-hand sides of the two equations are equal to 62, so they must be equal. Hence, 𝑥+𝑚𝐴𝐶𝐷=𝑥+𝑚𝐴𝐵𝐷+1.

Subtracting 𝑥 from both sides of the equation then yields 𝑚𝐴𝐶𝐷=𝑚𝐴𝐵𝐷+1.

Thus, 𝑚𝐴𝐵𝐷 is 1 smaller than 𝑚𝐴𝐶𝐷, so 𝑚𝐴𝐵𝐷<𝑚𝐴𝐶𝐷.

In our final example, we will use the properties of transversals of parallel lines to compare the measures of two angles in a given diagram.

Example 5: Proving an Inequality in a Triangle

Use <, =, or > to fill in the blank: 𝑚𝐴𝐶𝐷𝑚𝐴𝐵𝐶.

Answer

Let’s first mark the two angles we want to compare on the diagram.

It appears the angle at 𝐶 is larger than the angle at 𝐵; however, we want to prove this result. We can do this by noting that we are given a pair of parallel lines and 𝐵𝐶 is a transversal of these parallel lines. We then recall that alternate interior angles in a transversal of parallel lines are congruent. So, 𝑚𝐴𝐵𝐶=𝑚𝐵𝐶𝐷.

We know that 𝑚𝐴𝐶𝐷=𝑚𝐴𝐶𝐵+𝑚𝐵𝐶𝐷. Substituting 𝑚𝐵𝐶𝐷=𝑚𝐴𝐵𝐶 gives 𝑚𝐴𝐶𝐷=𝑚𝐴𝐶𝐵+𝑚𝐴𝐵𝐶.

Finally, since 𝑚𝐴𝐶𝐵>0, we can say that 𝑚𝐴𝐶𝐷>𝑚𝐴𝐵𝐶.

Let’s finish by recapping some of the important points from this explainer.

Key Points

  • The axioms of inequalities tell us for any real numbers 𝑎, 𝑏, 𝑐, and 𝑑, we have the following:
    • We can compare the sizes of any two real numbers. In other words, either 𝑎>𝑏, 𝑎<𝑏, or 𝑎=𝑏.
    • Inequalities and equalities are transitive relations on the set of real numbers. This means that if 𝑎>𝑏 and 𝑏>𝑐, then we must have that 𝑎>𝑐. This is called the transitivity axiom.
    • Adding a constant to or subtracting a constant from both sides of an inequality gives an equivalent inequality. For example, if 𝑎>𝑏, then 𝑎+𝑐>𝑏+𝑐 and 𝑎𝑐>𝑏𝑐.
    • Multiplying or dividing both sides of an inequality by a positive constant yields an equivalent inequality. For example, if 𝑎>𝑏 and 𝑐>0, then 𝑎𝑐>𝑏𝑐 and 𝑎𝑐>𝑏𝑐.
    • We can combine the ideas of the transitive axiom with the other axioms. For example, if 𝑎>𝑏 and 𝑐>𝑑, then 𝑎+𝑐>𝑏+𝑑.
  • The measure of any exterior angle in a triangle 𝐴𝐵𝐶 is greater than the measure of either nonadjacent interior angle in the triangle.

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