Lesson Explainer: Inequality in One Triangle: Side Comparison | Nagwa Lesson Explainer: Inequality in One Triangle: Side Comparison | Nagwa

Lesson Explainer: Inequality in One Triangle: Side Comparison Mathematics • Second Year of Preparatory School

In this explainer, we will learn how to form inequalities involving the lengths of the sides of a triangle given the measures of the angles of the triangle.

In an isosceles triangle, we can recall that the angles opposite the sides of equal lengths are of equal measures. The converse to this statement is also true: if two angles in a triangle have equal measures, then the sides opposite these angles have the same lengths. We can use this idea to consider what happens if we have two angles in a triangle with different measures.

We can start by considering an isosceles triangle. We know that the angles opposite the sides of equal lengths have equal measures. We can compare the lengths and measures of the angles by altering the base of the triangle. If we decrease one of the angles at the base (while leaving the third angle unchanged), then the other angle at the base must increase since the sum of the measures of the interior angles is constant at 180.

We can also see that doing this decreases the length of the side opposite the smaller angle. In other words, the smaller angle is opposite the smaller side. We can follow the same process if we increase the measures of one of the angles at the base.

Since we have increased the measure of one angle, the other angle must decrease in measure. We can also see that we have increased the length of the side opposite the angle of the larger measure.

This result holds true in general and is called the side comparison theorem in triangles.

Theorem: Side Comparison Theorem in Triangles

If we have a triangle such that two angles have unequal measures, then the side opposite the larger angle is longer than the side opposite the smaller angle. In particular, consider 𝐴𝐵𝐶.

If we know that 𝑚𝐵>𝑚𝐴, then 𝑏>𝑎.

It is also worth noting that we can apply this result with the inequality reversed. In particular, if we have a triangle such that two sides have unequal lengths, then the side opposite the smaller angle is shorter than the side opposite the angle with the larger measure.

Let’s see an example of applying this result to construct an inequality of the side lengths of a triangle using its angle measures.

Example 1: The Relation between Sides and Corresponding Angles in a Triangle

Arrange the side lengths of 𝐴𝐵𝐶 from least to greatest.

Answer

We are given two of the measures of the interior angles in a triangle and asked to use this to arrange the side lengths in ascending order. We can do this by recalling that the side comparison theorem for triangles tells us that, if we have a triangle such that two angles have unequal measures, then the side opposite the larger angle is longer than the side opposite the smaller angle.

We know that 94>59, so we can conclude that the side opposite 𝐵 is longer than the side opposite 𝐶. Thus, 𝐴𝐶>𝐴𝐵.

We can find the measure of 𝐴 by recalling that the sum of the measures of the interior angles in a triangle is 180. Therefore, 𝑚𝐴+𝑚𝐵+𝑚𝐶=180.

We can substitute the measures of 𝐵 and 𝐶 to get 𝑚𝐴+59+94=180.

We then solve for 𝑚𝐴: 𝑚𝐴=1809459𝑚𝐴=27.

We can now see that 𝑚𝐶>𝑚𝐴, so the side opposite 𝐶 must be longer than the side opposite 𝐴. Thus, 𝐴𝐵>𝐵𝐶.

We can combine these two inequalities into a compound inequality: 𝐴𝐶>𝐴𝐵>𝐵𝐶.

Finally, we are told to write the side lengths from least to greatest. This means we start with the shortest side and end with the longest side.

We can reverse the compound inequality to get the sides listed from least to greatest in length: 𝐵𝐶,𝐴𝐵,𝐴𝐶.

In our next example, we will use the side comparison theorem in triangles to determine which of two sides in a triangle is longer by using the measures of two exterior angles in the triangle.

Example 2: Completing Inequalities for the Side Lengths of a Triangle

From the figure, how do 𝐴𝐵 and 𝐵𝐶 compare?

Answer

We want to compare the lengths of two sides of triangle 𝐴𝐵𝐶. We can do this by recalling that the side comparison theorem in triangles tells us that, if we have a triangle such that two angles have unequal measures, then the side opposite the larger angle is longer than the side opposite the smaller angle.

This means we can compare the lengths of the sides of the triangle by comparing the measures of the angles opposite the sides. We can mark the angles opposite each side in the diagram.

We can see that 𝐴𝐶𝐵 and 𝐴𝐶𝐷 combine to make a straight angle, so their measures sum to give 180. Thus, 𝑚𝐴𝐶𝐵+𝑚𝐴𝐶𝐷=180.

We can see in the diagram that 𝑚𝐴𝐶𝐷=108, so we can substitute this value into the equation to get 𝑚𝐴𝐶𝐵+108=180.

We can then solve for 𝑚𝐴𝐶𝐵𝑚𝐴𝐶𝐵=180108=72.

We can now directly find 𝑚𝐵𝐴𝐶 by noting that 𝐶𝐵𝐸 is the opposite exterior angle in 𝐴𝐵𝐶 to 𝐵𝐴𝐶 and 𝐴𝐶𝐵.

We can recall that the measure of an exterior angle in a triangle is equal to the sum of the measures of the opposite interior angles. Thus, 𝑚𝐶𝐵𝐸=𝑚𝐵𝐴𝐶+𝑚𝐴𝐶𝐵.

We know that 𝑚𝐶𝐵𝐸=128 and 𝑚𝐴𝐶𝐵=72, so we can substitute these values into the equation to get 128=𝑚𝐵𝐴𝐶+72.

We can subtract 72 from both sides of the equation to get 𝑚𝐵𝐴𝐶=12872=56.

We can add this onto the diagram.

We see that the angle opposite 𝐴𝐵 has a larger measure than the angle opposite 𝐵𝐶; therefore, it must be the longer side.

Hence, 𝐴𝐵>𝐵𝐶.

In our next example, we will compare four pairs of side lengths in triangles by using the side comparison theorem in triangles.

Example 3: Completing Inequalities for the Side Lengths of a Triangle

Consider the given diagram.

Fill in the blanks in the following statements using >,<, or =.

  1. 𝐴𝐶𝐴𝐵
  2. 𝐴𝐵𝐵𝐶
  3. 𝐷𝐶𝐴𝐷
  4. 𝐴𝐷𝐴𝐶

Answer

We first recall the side comparison theorem in triangles that tells us that if we have a triangle such that two angles have unequal measures, then the side opposite the larger angle is longer than the side opposite the smaller angle.

This means we can compare the lengths of the sides of the triangle by comparing the measures of the angles opposite the sides.

Part 1

We want to compare the lengths of 𝐴𝐶 and 𝐴𝐵. We can start by highlighting these line segments and the angles opposite them on the diagram.

We can note that 90>45, so the side opposite the right angle must be longer than the side opposite the angle of measure 45.

Hence, 𝐴𝐶>𝐴𝐵.

Part 2

We can follow the same process to compare the lengths of 𝐴𝐵 and 𝐵𝐶. We highlight both line segments on the diagram as well as the angles opposite each side.

We can determine the measure of the angle opposite 𝐵𝐶 by adding the measures of the two angles that combine to form 𝐵𝐴𝐶. We have 𝑚𝐵𝐴𝐶=𝑚𝐵𝐴𝐷+𝑚𝐷𝐴𝐶=20+25=45.

Since both angles have the same measures, we can conclude that 𝐴𝐵𝐶 is isosceles with 𝐴𝐵=𝐵𝐶.

Part 3

To compare the lengths of 𝐷𝐶 and 𝐴𝐷, we can first highlight both line segments on the diagram as well as the angles opposite each side.

Since the measure of the angle opposite 𝐷𝐶 is smaller than the measure of the angle opposite 𝐴𝐷, we can conclude that 𝐷𝐶 is shorter than 𝐴𝐷.

Hence, 𝐷𝐶<𝐴𝐷.

Part 4

Finally, to compare the lengths of 𝐴𝐷 and 𝐴𝐶, we highlight both line segments on the diagram as well as the angles opposite each side.

It appears as though the measure of 𝐴𝐷𝐶 is greater than that of 𝐴𝐶𝐷. However, for due diligence, we should verify this. We can find the measure of this angle by using the fact that the sum of the measures of the interior angles in a triangle is 180. Applying this to 𝐴𝐶𝐷, we have 𝑚𝐴𝐶𝐷+𝑚𝐶𝐴𝐷+𝑚𝐴𝐷𝐶=180.

Substituting the known angle measures into the equation gives 45+25+𝑚𝐴𝐷𝐶=180.

We can then solve for 𝑚𝐴𝐷𝐶: 𝑚𝐴𝐷𝐶=1804525=110.

We can now see that 𝑚𝐴𝐷𝐶>𝑚𝐴𝐶𝐷, so the side opposite 𝐴𝐷𝐶 is longer than the side opposite 𝐴𝐶𝐷.

Hence, 𝐴𝐷<𝐴𝐶.

In the previous example, we saw that the side opposite the right angle in a right triangle was longer than the other side. We can show that this result is true in general.

We know that the sum of the measures of the interior angles in a triangle is 180. So, in this triangle, we have 𝑚𝐴+𝑚𝐵+𝑚𝐶=180.

Substituting 𝑚𝐵=90 into the equation and rearranging gives us 𝑚𝐴+90+𝑚𝐶=180𝑚𝐴+𝑚𝐶=90.

Since the angle measures must be positive, we can note that both of these angles have measures less than 90. In particular, this tells us that 𝐵 has the largest measure in the triangle. Therefore, the side opposite the right angle in a right triangle is the longest side. We call this the hypotenuse of the right triangle.

This agrees with our knowledge of the Pythagorean theorem. We recall that this tells us that, in a right triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the two other sides. In our right triangle above, this is 𝑏=𝑎+𝑐.

Since 𝑎,𝑏, and 𝑐 are all lengths, they are all positive, so we can note that 𝑏>𝑎 and 𝑏>𝑐. Thus, the hypotenuse must be the longest side.

In our next example, we will prove an inequality involving the lengths of lines in a geometric construct.

Example 4: Proving an Inequality Involving the Side Lengths of a Triangle

From the figure, fill in the blank with either >,<, or =: 𝐴𝐶𝐵𝐶.

Answer

We want to compare the lengths of two sides of a triangle. We can do this by recalling that the longer side will be opposite the angle with the larger measure. Let’s start by highlighting the sides in the given inequality and the angles opposite each side. We will call the angle measures 𝑥 and 𝑦 as shown.

We can see that the angles at 𝐴 make a straight angle, so their measures sum to give 180. Thus, 𝑥+50+70=180.

Solving for 𝑥 gives 𝑥=1805070=60.

We can find 𝑦 by noting that it is an alternate interior angle to another angle in a transversal of parallel lines.

We can then find 𝑦 by using the fact that the sum of the measures in a straight angle is 180 or by noting that 𝑦 is the corresponding angle of the angle of measure 70 in the diagram.

Since the angle opposite 𝐴𝐶 has a larger measure than the angle opposite 𝐵𝐶, we can conclude that 𝐴𝐶 is longer than 𝐵𝐶. Hence, 𝐴𝐶>𝐵𝐶.

We can prove the side comparison theorem by considering a triangle in which 𝑚𝐴>𝑚𝐵.

We can first show that 𝐵𝐶 and 𝐴𝐶 do not have the same length; if they did, then 𝐴𝐵𝐶 would be an isosceles triangle since it has two sides of equal lengths. We can recall that the angles opposite the sides of equal lengths must have equal measures. This would mean that 𝑚𝐴=𝑚𝐵, but we are told that 𝑚𝐴>𝑚𝐵, so this cannot be true. Hence, 𝐵𝐶 and 𝐴𝐶 must have unequal lengths.

We can also show that 𝐴𝐶 cannot be longer than 𝐵𝐶. We recall that the angle comparison theorem in triangles tells us that if we have a triangle such that two sides have unequal lengths, then the angle opposite the longer side has a larger measure than the angle opposite the shorter side.

If we assume that 𝐴𝐶 is longer than 𝐵𝐶, then this theorem tells us that the angle opposite 𝐴𝐶 must have a larger measure than the angle opposite 𝐵𝐶. In other words, we would have 𝑚𝐵>𝑚𝐴. However, we are told that 𝑚𝐴>𝑚𝐵, so this cannot be the case. Hence, 𝐴𝐶 cannot be longer than 𝐵𝐶.

Since 𝐵𝐶 and 𝐴𝐶 cannot have the same length and 𝐴𝐶 cannot be longer than 𝐵𝐶, there is only one final possibility; 𝐵𝐶 is longer than 𝐴𝐶.

In our next example, we will use the side comparison theorem to compare lengths in a given construct.

Example 5: Comparing Side lengths in a Triangle Using the Side Comparison Theorem

Consider the figure shown.

Fill in the blank with >,<, or =: 𝐴𝐵𝐵𝐶.

Answer

From the diagram, we can note that 𝐴𝐷=𝐷𝐶, so triangle 𝐴𝐷𝐶 is an isosceles triangle. The angles opposite the sides of equal lengths must be congruent. We can add these congruent angles onto the diagram.

We can now recall that the side comparison theorem in triangles tells us that if we have a triangle such that two angles have unequal measures, then the side opposite the larger angle is longer than the side opposite the smaller angle.

We see that 𝐵𝐴𝐶 is opposite 𝐵𝐶 and 𝐴𝐶𝐵 is opposite 𝐴𝐵. We can also see from the diagram that 𝐵𝐴𝐶 has a larger measure than 𝐴𝐶𝐵 since it is the combination of two angles, one of which has an equal measure to 𝐴𝐶𝐵: 𝑚𝐵𝐴𝐶=𝑚𝐵𝐴𝐷+𝑚𝐷𝐴𝐶=𝑚𝐵𝐴𝐷+𝑚𝐴𝐶𝐵.

We can conclude that the side opposite the angle with a lower measure is shorter. Hence, 𝐴𝐵<𝐵𝐶.

We can use the angle comparison theorem in triangles to prove many useful geometric results. For example, consider 𝐴𝐵𝐶 and line segment 𝐴𝐷 such that 𝐴𝐷𝐵𝐶.

We can show that 𝐴𝐷 is the shortest line segment from 𝐴 to 𝐵𝐶. Consider any other point 𝐸𝐵𝐶.

We see that 𝐴𝐷𝐸 is a right triangle, so its hypotenuse is the longest side. Thus, 𝐴𝐸>𝐴𝐷.

This is true for any point 𝐸 on 𝐵𝐶, so 𝐴𝐷 is the shortest line segment from 𝐴 to 𝐵𝐶.

This holds for any perpendicular line from a point to a straight line.

Property: Perpendicular from a Point to a Line

The perpendicular line segment connecting a line and a point is the shortest line segment connecting that line and that point.

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

Key Points

  • If we have a triangle such that two angles have unequal measures, then the side opposite the larger angle is longer than the side opposite the smaller angle.
  • In a right triangle, the hypotenuse is the longest side since it is opposite the right angle, which is the angle with the largest measure.
  • The perpendicular line segment drawn from one side of a triangle to the opposite vertex is shorter than any other line segment drawn from the same side to the same vertex.

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