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Lesson Explainer: Similarity of Triangles Mathematics • 8th Grade

In this explainer, we will learn how to determine and prove whether two triangles are similar using equality of corresponding angles or proportionality of corresponding sides and how to use similarity to find unknown lengths and angles.

Let’s begin with the definition of similar triangles.

Definition: Similar Triangles

Similar triangles have corresponding angles congruent and corresponding sides proportional.

We can prove that two triangles are similar if

  • corresponding angles are congruent or
  • corresponding sides are proportional.

Note that the definition of congruent triangles is different since in congruent triangles, corresponding angles are congruent and corresponding sides are congruent.

We can see an example of similarity with the triangles 𝐴𝐡𝐢 and 𝐷𝐸𝐹 below.

There are 3 pairs of corresponding angle measures congruent: π‘šβˆ π΄=π‘šβˆ π·,π‘šβˆ π΅=π‘šβˆ πΈ,π‘šβˆ πΆ=π‘šβˆ πΉ.

We can also write that the sides are in proportion since we have 𝐴𝐡𝐷𝐸=𝐴𝐢𝐷𝐹=𝐡𝐢𝐸𝐹714=36=510ο€Ό=12.

It would also be valid to write the proportion of the sides in the triangles with all the sides of △𝐷𝐸𝐹 as the numerators and all the sides of △𝐴𝐡𝐢 as the denominators (that is, the fractions have all been flipped) such that 𝐷𝐸𝐴𝐡=𝐷𝐹𝐴𝐢=𝐸𝐹𝐡𝐢147=63=105.

In this case, the ratios of the sides would all be equivalent to 2.

Either proportion statement would be enough to prove that β–³π΄π΅πΆβˆΌβ–³π·πΈπΉ.

When proving that two triangles are similar, demonstrating either that the angles are congruent or that the sides are in proportion is enough to prove that the triangles are similar. This is because if we are given 3 angle measures, we can only draw another similar triangle with the same angle measures and not a dissimilar triangle, as all the sides would be proportional to the original triangle. Or, given 3 side lengths, if we draw another triangle with sides in the same proportion, all the angle measures would be congruent to those of the original triangle.

The notation that we use when writing similarity relationships is important. Similar shapes can be related by the symbol ∼. However, the order in which we write the vertices of the shapes is very important because the similarity relationship itself indicates the vertices (and sides) that are corresponding.

For example, in the figure above, we could write that β–³π΄π΅πΆβˆΌβ–³π·πΈπΉ. If we wrote, for example, that β–³π΄π΅πΆβˆΌβ–³πΈπ·πΉ, this would be incorrect since vertex 𝐴 corresponds to vertex 𝐷, not 𝐸. We could, however, write a number of different correct similarity statements, for example, β–³π΅πΆπ΄βˆΌβ–³πΈπΉπ·,β–³πΆπ΄π΅βˆΌβ–³πΉπ·πΈ,β–³π΅π΄πΆβˆΌβ–³πΈπ·πΉ.

In the first examples, we will see how we can prove that two triangles are similar, beginning with an example where we use the side lengths of the triangles.

Example 1: Identifying Whether Triangles Are Similar by Considering Their Sides

Is triangle 𝑀𝑁𝐿 similar to triangle π‘‹π‘π‘Œ?

Answer

Similar triangles have corresponding angles congruent and corresponding sides proportional. As one of these properties leads to the other, we can prove that triangles are similar if they either have corresponding angles congruent or corresponding sides proportional.

Given that we have the lengths of the sides in this figure, let’s determine the ratio between each of the corresponding side lengths.

We observe that both triangles are equilateral since each triangle has 3 congruent side lengths. Using corresponding pairs of sides, 𝑀𝑁=12cm and 𝑋𝑍=18cm, 𝑀𝐿=12cm and π‘‹π‘Œ=18cm, and 𝑁𝐿=12cm and π‘π‘Œ=18cm, we can write that 𝑀𝑁𝑋𝑍=π‘€πΏπ‘‹π‘Œ=π‘πΏπ‘π‘Œ since these are all equal to the ratio 1218, or 23.

Hence, we can answer that triangle 𝑀𝑁𝐿 is similar to triangle π‘‹π‘π‘Œ.

Note, we could also have demonstrated the triangles are similar by writing the ratios with all the numerators and denominators swapped. That is, 𝑋𝑍𝑀𝑁=π‘‹π‘Œπ‘€πΏ=π‘π‘Œπ‘πΏ.

In this case, these ratios would all still be equal but this time to 1812, or 32.

In the previous example, we used the fact that the sides are proportional to each other to prove that the triangles are similar. However, we could have used an alternative method. Having established that both triangles are equilateral, we could recall that equilateral triangles have all three angle measures equal to 60∘. Since the corresponding angle measures in each triangle are congruent, then the triangles are similar.

As an aside, we can note that all regular polygons will be similar. For example, all squares are similar, all regular pentagons are similar, all regular hexagons are similar, and so on. An equilateral triangle, being a regular triangle, is always similar to any other equilateral triangle.

We will now see another example where we will be using angle measures to establish a pair of similar triangles.

Example 2: Identifying Whether Triangles Are Similar by Considering Their Angles

Which of the following triangles is similar to the one seen in the given figure?

Answer

We recall that two triangles are similar if they have corresponding angle measures congruent and corresponding sides proportional. One way we can prove that triangles are similar is if they have corresponding angle measures congruent.

When we consider the 5 different choices, we can see that none of the available triangles has angles that have the same measure as the given triangle, 100∘ and 30∘. Therefore, it will be useful to calculate the measure of the third angle in this triangle, which we can define as π‘₯∘. Using the property that the sum of the internal angle measures in a triangle is 180∘, we have that π‘₯+100+30=180π‘₯+130=180π‘₯=180βˆ’130=50.∘∘∘∘∘∘∘∘∘∘∘

If we look at the available choices, we see that the only triangle that has two congruent angle measures is choice E. We can calculate the missing angle, defined as π‘¦βˆ˜, in choice E using the fact that the sum of the angle measures in the triangle is 180∘: 𝑦+50+30=180𝑦+80=180𝑦=180βˆ’80=100.∘∘∘∘∘∘∘∘∘∘∘

Thus, all 3 corresponding angle measures of the given triangle are congruent with those in choice E.

Therefore, the triangle that is similar to the given figure is the triangle in choice E.

In the previous example, we calculated the measure of the third angle in the given figure and then calculated the measure of the third angle in choice E to establish that its measure is 100∘. However, if we know that two pairs of corresponding angles in two triangles are congruent, then the third pair of angles in the triangles must also be congruent. This arises directly from the fact that the interior angle measures in a triangle sum to 180∘.

In the figure below, if we are given two pairs of congruent angle measures, π‘Žβˆ˜ and π‘βˆ˜, then the third angle, π‘βˆ˜, in each triangle would be equal to 𝑐=180βˆ’(π‘Ž+𝑏)∘∘∘∘.

We will now see another example.

Example 3: Identifying Similar Triangles Using the Angle Property of Isosceles Triangles

Which two of these triangles are similar?

Answer

We recall that two triangles are similar if they have corresponding angles congruent and corresponding sides proportional. We can prove two triangles are similar either by determining if corresponding angles are congruent or by determining if corresponding sides are proportional.

In this question, we are not given any information about the side lengths of these triangles. So, let us see if we can calculate the angle measures in the triangles. We can observe that all 4 triangles must be isosceles, as each triangle has a pair of congruent sides marked. In an isosceles triangle, the two base angles are of equal measure. We also know that the internal angle measures in a triangle sum to 180∘.

Beginning with triangle 1, we know that since a base angle is 40∘, the other base angle is also 40∘. Subtracting these from 180∘, we can find the measure of the third angle, defined as π‘Žβˆ˜, as π‘Ž+40+40=180π‘Ž+80=180π‘Ž=180βˆ’80=100.∘∘∘∘∘∘∘∘∘∘∘

We could continue to calculate all the missing angles in the other figures; however, it is useful to observe that the only other triangle given that also has a vertex angle of 100∘ is triangle 4.

Defining the 2 congruent base angles in triangle 4 as π‘βˆ˜, we could calculate their measure, using the vertex angle of 100∘, as 𝑏+𝑏+100=1802𝑏+100=1802𝑏=180βˆ’1002𝑏=80𝑏=802=40.∘∘∘∘∘∘∘∘∘∘∘∘∘∘∘

Thus, these two triangles are similar.

For completeness, we could establish all the missing angles in each triangle as below.

Even without calculating these angles, we can observe that triangles 2 and 3 have two noncongruent vertex angle measures of 40∘ and 90∘; therefore, they will not be similar to each other or to triangles 1 and 4.

Hence, the two triangles that are similar are 1 and 4.

In the next example, we will see how we can find an unknown side length by first establishing if two triangles are similar.

Example 4: Finding a Missing Length Using Similarity

Determine the length of 𝐴𝐢.

Answer

In the given figure, we have two triangles of different side lengths. This means that the triangles are not congruent. However, we can check if they are similar. We recall that two triangles are similar if they have corresponding angles congruent and corresponding sides proportional. We can prove two triangles are similar either by determining if corresponding angles are congruent or by determining if corresponding sides are proportional.

We do not have enough information to compare all the side lengths, so we check the angle measures. As we have 2 angle measures given in each triangle, we can use the property that the internal angle measures in a triangle sum to 180∘ to help us calculate the third angle in each triangle.

In β–³π‘‹π‘Œπ‘, we are given that π‘šβˆ π‘‹π‘π‘Œ=61∘ and π‘šβˆ π‘‹π‘Œπ‘=60∘; thus, we can calculate π‘šβˆ π‘π‘‹π‘Œ as π‘šβˆ π‘‹π‘π‘Œ+π‘šβˆ π‘‹π‘Œπ‘+π‘šβˆ π‘π‘‹π‘Œ=18061+60+π‘šβˆ π‘π‘‹π‘Œ=180121+π‘šβˆ π‘π‘‹π‘Œ=180π‘šβˆ π‘π‘‹π‘Œ=180βˆ’121=59.∘∘∘∘∘∘∘∘∘

As we have now established that 61+60+59=180∘∘∘∘, then π‘šβˆ π΄π΅πΆ can be determined to be 60∘.

We now have that π‘šβˆ π‘‹=π‘šβˆ π΄,π‘šβˆ π‘Œ=π‘šβˆ π΅,π‘šβˆ π‘=π‘šβˆ πΆ.

As there are 3 pairs of corresponding angle measures congruent, we have proven that β–³π‘‹π‘Œπ‘βˆΌβ–³π΄π΅πΆ.

We can then use this information to determine the length of 𝐴𝐢. Side 𝐴𝐢 in △𝐴𝐡𝐢 corresponds to side 𝑋𝑍 in β–³π‘‹π‘Œπ‘.

In order to find the length of 𝐴𝐢, we look for another pair of corresponding sides for which we are given the length measurements. We observe that we are given that the corresponding sides 𝐡𝐢 and π‘Œπ‘ are equal to 22.8 cm and 12 cm respectively.

So, we can write a proportion statement and then substitute the length values. This gives us 𝐴𝐢𝑋𝑍=π΅πΆπ‘Œπ‘π΄πΆ12.1=22.812.

Multiplying both sides by 12.1, we have 𝐴𝐢=12.1Γ—22.812=22.99.cm

Therefore, we can give the answer that the length of 𝐴𝐢 is 22.99 cm.

In the previous example, we first proved that two triangles are similar and used this to find the length of a side. In the next example, we will move beyond this to finding side lengths in order to perform a further calculation: finding the perimeter of a triangle. Recall that the perimeter of a polygon is the distance around its outside edge.

To do this, we will need to understand the similarity ratio (often called the scale factor) between two similar shapes. Consider the similar triangles 𝐴𝐡𝐢 and 𝐴′𝐡′𝐢′ below.

We can confirm that β–³π΄π΅πΆβˆΌβ–³π΄β€²π΅β€²πΆβ€² since the corresponding side lengths are all in the same proportion. That is, we can write that 𝐴′𝐡′𝐴𝐡=𝐡′𝐢′𝐡𝐢=𝐢′𝐴′𝐢𝐴.

By substituting in the lengths of any two corresponding sides, we can establish the similarity ratio. Given that 𝐴𝐡=4cm and 𝐴′𝐡′=8cm, we have that the similarity ratio from △𝐴𝐡𝐢 to △𝐴′𝐡′𝐢′ can be determined as similarityratio=𝐴′𝐡′𝐴𝐡=84=2.

Furthermore, since perimeter is also a measure of length, then the perimeters of two similar triangles (and indeed any two similar polygons) will be in the same ratio as the similarity ratio between them.

We can demonstrate this by calculating the perimeters of △𝐴𝐡𝐢 and △𝐴′𝐡′𝐢′ in the figure above as follows: perimeterofcmperimeterofcm△𝐴𝐡𝐢=4+6+5=15,△𝐴′𝐡′𝐢′=8+12+10=30.

We can then write the ratio of the perimeters as perimeterofperimeterof△𝐴′𝐡′𝐢′△𝐴𝐡𝐢=3015=2.

As we know that the similarity ratio from △𝐴𝐡𝐢 to △𝐴′𝐡′𝐢′ was also 2, then we have confirmed that perimeterofperimeterofsimilarityratio△𝐴′𝐡′𝐢′△𝐴𝐡𝐢=.

Let’s now see how this can be applied in the following example.

Example 5: Finding the Perimeter of a Triangle Using the Similarity between Two Triangles

𝐴𝐡𝐢𝐷 is a rectangle in which 𝐴𝐷=21cm, 𝐴𝑋=9cm, and 𝑋𝑀=12cm. Calculate the perimeter of β–³π‘Œπ‘€πΆ.

Answer

It is often useful to begin a question such as this by writing any given length measurements on the diagram and establishing exactly what we are asked to calculate.

We are given that 𝐴𝐡𝐢𝐷 is a rectangle, so opposite sides are parallel and congruent. From the diagram, we have that π‘‹π‘Œβˆ₯𝐴𝐡, so we know that all 3 vertical line segments are parallel: π‘‹π‘Œβˆ₯𝐴𝐡βˆ₯𝐷𝐢. As 𝐴𝐡𝐢𝐷 is a rectangle, then all 3 vertical line segments are perpendicular to 𝐴𝐷 and 𝐡𝐢.

We will consider if we have similar triangles in this figure. We recall that two triangles are similar if they have corresponding angles congruent and corresponding sides in the same proportion. One way we can prove two triangles are similar is by demonstrating that corresponding angles are congruent. Let’s see if we can use the parallel lines to determine any further congruent angle measures.

Using the transversal 𝐴𝐢, we have that βˆ π‘‹π΄π‘€ is alternate to βˆ π‘ŒπΆπ‘€; hence, π‘šβˆ π‘‹π΄π‘€=π‘šβˆ π‘ŒπΆπ‘€.

Furthermore, we also have a pair of vertically opposite angles: βˆ π‘‹π‘€π΄ and βˆ π‘Œπ‘€πΆ; hence, π‘šβˆ π‘‹π‘€π΄=π‘šβˆ π‘Œπ‘€πΆ.

As π‘‹π‘Œβˆ₯𝐷𝐢 and by the fact that we know the rectangle has a right angle at ∠𝐴𝐷𝐢, the corresponding angle at βˆ π΄π‘‹π‘€ will be congruent. Similarly, π‘šβˆ πΆπ΅π΄=π‘šβˆ πΆπ‘Œπ‘€=90∘. Thus we have a third pair of corresponding angles in the triangles: π‘šβˆ π΄π‘‹π‘€=π‘šβˆ πΆπ‘Œπ‘€(=90).∘

Therefore, since we have 3 pairs of congruent angles, we can write that β–³π‘‹π‘€π΄βˆΌβ–³π‘Œπ‘€πΆ.

We can use the similarity of these triangles to find the perimeter of β–³π‘Œπ‘€πΆ. The ratio between the perimeters of two similar triangles is equal to the ratio between any two corresponding sides (the similarity ratio). As the perimeter is the distance around the outside edge, we can calculate the perimeter of △𝑋𝑀𝐴 and then apply the similarity ratio from △𝑋𝑀𝐴 to β–³π‘Œπ‘€πΆ to determine the perimeter of β–³π‘Œπ‘€πΆ.

We note that we have one unknown side length, 𝐴𝑀, in △𝑋𝑀𝐴. However, given that this is a right triangle, we can apply the Pythagorean theorem. The Pythagorean theorem states that in a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

Thus, given 𝑋𝑀=12cm and 𝐴𝑋=9cm, the length of the hypotenuse 𝐴𝑀 can be calculated as (𝐴𝑀)=(𝑋𝑀)+(𝐴𝑋)(𝐴𝑀)=12+9(𝐴𝑀)=144+81(𝐴𝑀)=225.

We can then take the square root of both sides, and since 𝐴𝑀 is a length, we consider only the positive value. Hence, 𝐴𝑀=√225=15.cm

The perimeter of △𝑋𝑀𝐴 can then be calculated, given 𝐴𝑋=9cm, 𝑋𝑀=12cm, and 𝐴𝑀=15cm, as perimeterofcm△𝑋𝑀𝐴=𝐴𝑋+𝑋𝑀+𝐴𝑀=9+12+15=36.

Next, we determine the similarity ratio between the two triangles by identifying the lengths of a pair of corresponding side lengths. Although it does not appear that we have any measurements in β–³π‘Œπ‘€πΆ, we use the property that in a rectangle opposite sides are congruent; thus, 𝐡𝐢=𝐴𝐷=21cm. Furthermore, since π΄π‘‹π‘Œπ΅ also forms a rectangle, then π΅π‘Œ=𝐴𝑋=9cm. Hence, we can calculate the length of πΆπ‘Œ as πΆπ‘Œ=π΅πΆβˆ’π΅π‘Œ=21βˆ’9=12.cm

Note that in the similar triangles 𝑋𝑀𝐴 and π‘Œπ‘€πΆ, sides πΆπ‘Œ (12 cm) and 𝐴𝑋 (9 cm) are corresponding.

We can determine the similarity ratio from △𝑋𝑀𝐴 to β–³π‘Œπ‘€πΆ using these corresponding side lengths as similarityratiofromtoβ–³π‘‹π‘€π΄β–³π‘Œπ‘€πΆ=πΆπ‘Œπ΄π‘‹=129=43.

As previously mentioned, the ratio between the perimeters of two similar triangles is equal to the ratio between any two corresponding sides. Given that the perimeter of △𝑋𝑀𝐴 is 36 cm, we multiply this by the similarity ratio from △𝑋𝑀𝐴 to β–³π‘Œπ‘€πΆ. This gives perimeterofcmβ–³π‘Œπ‘€πΆ=36Γ—43=1443=48.

Thus, by first proving that the two triangles are similar and applying the similarity ratio between triangles, we determined that the perimeter of β–³π‘Œπ‘€πΆ is 48 cm.

In the previous example, we saw how there was a pair of similar triangles created by parallel lines and a transversal within the rectangle. In general, we always have similar triangles created by the following two geometric arrangements involving parallel lines since we can prove that alternate, corresponding, and vertically opposite angles are congruent.

Thus, when solving problems involving similar triangles, it is very important to be able to use and recall a wide variety of angle properties, such as those in parallel lines, vertically opposite angles, the angle sum on a straight line, and the sum of the angle measures in a triangle. Depending on the problem at hand, some of these properties may allow us to prove that two triangles are similar.

We will now summarize the key points.

Key Points

  • Similar triangles have corresponding angles congruent and corresponding sides porportional.
  • We can prove that two triangles are similar if
    • corresponding angles are congruent or
    • corresponding sides are porportional.
  • When writing a similarity relationship between two triangles, the order of the vertices is important. Corresponding vertices should be in the same position in the similarity statement.
  • The ratio between the perimeters of two similar triangles is equal to the ratio between any two corresponding sides.

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