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

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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 △𝐴𝐡𝐢.

If we know that π‘šβˆ π΅>π‘šβˆ π΄, then 𝑏>π‘Ž.

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 △𝐴𝐡𝐢 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 ∠𝐡 is longer than the side opposite ∠𝐢. Thus, 𝐴𝐢>𝐴𝐡.

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, π‘šβˆ π΄+π‘šβˆ π΅+π‘šβˆ πΆ=180.∘

We can substitute the measures of ∠𝐡 and ∠𝐢 to get π‘šβˆ π΄+59+94=180.∘∘∘

We then solve for π‘šβˆ π΄: π‘šβˆ π΄=180βˆ’94βˆ’59π‘šβˆ π΄=27.∘∘∘∘

We can now see that π‘šβˆ πΆ>π‘šβˆ π΄, so the side opposite ∠𝐢 must be longer than the side opposite ∠𝐴. Thus, 𝐴𝐡>𝐡𝐢.

We can combine these two inequalities into a compound inequality: 𝐴𝐢>𝐴𝐡>𝐡𝐢.

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: 𝐡𝐢,𝐴𝐡,𝐴𝐢.

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 𝐴𝐡 and 𝐡𝐢 compare?

Answer

We want to compare the lengths of two sides of triangle △𝐴𝐡𝐢. 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 ∠𝐴𝐢𝐡 and ∠𝐴𝐢𝐷 combine to make a straight angle, so their measures sum to give 180∘. Thus, π‘šβˆ π΄πΆπ΅+π‘šβˆ π΄πΆπ·=180.∘

We can see in the diagram that π‘šβˆ π΄πΆπ·=108∘, so we can substitute this value into the equation to get π‘šβˆ π΄πΆπ΅+108=180.∘∘

We can then solve for π‘šβˆ π΄πΆπ΅π‘šβˆ π΄πΆπ΅=180βˆ’108=72.∘∘∘

We can now directly find π‘šβˆ π΅π΄πΆ by noting that ∠𝐢𝐡𝐸 is the opposite exterior angle in △𝐴𝐡𝐢 to ∠𝐡𝐴𝐢 and ∠𝐴𝐢𝐡.

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, π‘šβˆ πΆπ΅πΈ=π‘šβˆ π΅π΄πΆ+π‘šβˆ π΄πΆπ΅.

We know that π‘šβˆ πΆπ΅πΈ=128∘ and π‘šβˆ π΄πΆπ΅=72∘, so we can substitute these values into the equation to get 128=π‘šβˆ π΅π΄πΆ+72.∘∘

We can subtract 72∘ from both sides of the equation to get π‘šβˆ π΅π΄πΆ=128βˆ’72=56.∘∘∘

We can add this onto the diagram.

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

Hence, 𝐴𝐡>𝐡𝐢.

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. 𝐴𝐢𝐴𝐡
  2. 𝐴𝐡𝐡𝐢
  3. 𝐷𝐢𝐴𝐷
  4. 𝐴𝐷𝐴𝐢

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 𝐴𝐢 and 𝐴𝐡. 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, 𝐴𝐢>𝐴𝐡.

Part 2

We can follow the same process to compare the lengths of 𝐴𝐡 and 𝐡𝐢. 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 𝐡𝐢 by adding the measures of the two angles that combine to form ∠𝐡𝐴𝐢. We have π‘šβˆ π΅π΄πΆ=π‘šβˆ π΅π΄π·+π‘šβˆ π·π΄πΆ=20+25=45.∘∘∘

Since both angles have the same measures, we can conclude that △𝐴𝐡𝐢 is isosceles with 𝐴𝐡=𝐡𝐢.

Part 3

To compare the lengths of 𝐷𝐢 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 𝐷𝐢 is smaller than the measure of the angle opposite 𝐴𝐷, we can conclude that 𝐷𝐢 is shorter than 𝐴𝐷.

Hence, 𝐷𝐢<𝐴𝐷.

Part 4

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

It appears as though the measure of ∠𝐴𝐷𝐢 is greater than that of ∠𝐴𝐢𝐷. 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 △𝐴𝐢𝐷, we have π‘šβˆ π΄πΆπ·+π‘šβˆ πΆπ΄π·+π‘šβˆ π΄π·πΆ=180.∘

Substituting the known angle measures into the equation gives 45+25+π‘šβˆ π΄π·πΆ=180.∘∘∘

We can then solve for π‘šβˆ π΄π·πΆ: π‘šβˆ π΄π·πΆ=180βˆ’45βˆ’25=110.∘∘∘∘

We can now see that π‘šβˆ π΄π·πΆ>π‘šβˆ π΄πΆπ·, so the side opposite ∠𝐴𝐷𝐢 is longer than the side opposite ∠𝐴𝐢𝐷.

Hence, 𝐴𝐷<𝐴𝐢.

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 π‘šβˆ π΄+π‘šβˆ π΅+π‘šβˆ πΆ=180.∘

Substituting π‘šβˆ π΅=90∘ into the equation and rearranging gives us π‘šβˆ π΄+90+π‘šβˆ πΆ=180π‘šβˆ π΄+π‘šβˆ πΆ=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 ∠𝐡 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=π‘Ž+𝑐.

Since π‘Ž,𝑏, and 𝑐 are all lengths, they are all positive, so we can note that 𝑏>π‘ŽοŠ¨οŠ¨ and 𝑏>π‘οŠ¨οŠ¨. 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 =: 𝐴𝐢𝐡𝐢.

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 π‘₯ and 𝑦 as shown.

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

Solving for π‘₯ gives π‘₯=180βˆ’50βˆ’70=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 𝐴𝐢 has a larger measure than the angle opposite 𝐡𝐢, we can conclude that 𝐴𝐢 is longer than 𝐡𝐢. Hence, 𝐴𝐢>𝐡𝐢.

We can prove the side comparison theorem by considering a triangle in which π‘šβˆ π΄>π‘šβˆ π΅.

We can first show that 𝐡𝐢 and 𝐴𝐢 do not have the same length; if they did, then △𝐴𝐡𝐢 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 π‘šβˆ π΄=π‘šβˆ π΅, but we are told that π‘šβˆ π΄>π‘šβˆ π΅, so this cannot be true. Hence, 𝐡𝐢 and 𝐴𝐢 must have unequal lengths.

We can also show that 𝐴𝐢 cannot be longer than 𝐡𝐢. 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 𝐴𝐢 is longer than 𝐡𝐢, then this theorem tells us that the angle opposite 𝐴𝐢 must have a larger measure than the angle opposite 𝐡𝐢. In other words, we would have π‘šβˆ π΅>π‘šβˆ π΄. However, we are told that π‘šβˆ π΄>π‘šβˆ π΅, so this cannot be the case. Hence, 𝐴𝐢 cannot be longer than 𝐡𝐢.

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

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 =: 𝐴𝐡𝐡𝐢.

Answer

From the diagram, we can note that 𝐴𝐷=𝐷𝐢, so triangle 𝐴𝐷𝐢 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 ∠𝐡𝐴𝐢 is opposite 𝐡𝐢 and ∠𝐴𝐢𝐡 is opposite 𝐴𝐡. We can also see from the diagram that ∠𝐡𝐴𝐢 has a larger measure than ∠𝐴𝐢𝐡 since it is the combination of two angles, one of which has an equal measure to ∠𝐴𝐢𝐡: π‘šβˆ π΅π΄πΆ=π‘šβˆ π΅π΄π·+π‘šβˆ π·π΄πΆ=π‘šβˆ π΅π΄π·+π‘šβˆ π΄πΆπ΅.

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

We can use the angle comparison theorem in triangles to prove many useful geometric results. For example, consider △𝐴𝐡𝐢 and line segment 𝐴𝐷 such that π΄π·βŸ‚βƒ–οƒ©οƒ©οƒ©οƒ©βƒ—π΅πΆ.

We can show that 𝐴𝐷 is the shortest line segment from 𝐴 to ⃖⃗𝐡𝐢. Consider any other point πΈβˆˆβƒ–οƒ©οƒ©οƒ©οƒ©βƒ—π΅πΆ.

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 ⃖⃗𝐡𝐢, so 𝐴𝐷 is the shortest line segment from 𝐴 to ⃖⃗𝐡𝐢.

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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