Lesson Explainer: Triangle Similarity Criteria and Their Applications | Nagwa Lesson Explainer: Triangle Similarity Criteria and Their Applications | Nagwa

Lesson Explainer: Triangle Similarity Criteria and Their Applications Mathematics • First Year of Secondary School

In this explainer, we will learn how to use the properties of similar triangles to solve problems.

We can begin by understanding the meaning of similar.

Definition: Similar Triangles

Two triangles are similar if corresponding angles are congruent and the lengths of their corresponding sides are in the same proportion.

More colloquially, we might say that similar triangles are the same shape, but they can be a different size. As an aside, triangles that are the same shape and the same size are defined as congruent.

Below is an example of two triangles that are similar.

The angle pairs, 𝐴 and 𝐷, 𝐵 and 𝐸, and 𝐶 and 𝐹, are of equal measures. The lengths of the corresponding sides, 𝐴𝐵 and 𝐷𝐸, 𝐴𝐶 and 𝐷𝐹, and 𝐵𝐶 and 𝐸𝐹, are in the same proportion. In this example, we can also say that the scale factor from 𝐴𝐵𝐶 to 𝐷𝐸𝐹 is 12.

We will now investigate the geometry of similar triangles. We can take a triangle, such as the triangle 𝐹𝐺𝐻.

Performing a dilation of scale factor 𝑘 would produce the following triangle 𝐹𝐺𝐻.

In a dilation, all side lengths are multiplied by the scale factor, and all the angle measures are preserved. If the scale factor is greater than 1, then the figure is enlarged. And if the scale factor is less than 1, then the figure is reduced.

We can say that triangle 𝐹𝐺𝐻 is similar to triangle 𝐹𝐺𝐻, and we can write this as 𝐹𝐺𝐻𝐹𝐺𝐻.

When we write the similarity relationship between triangles, the ordering of the lettering is important, as it indicates the angles and sides that correspond in the triangles.

If a given pair of triangles have equal corresponding angles, then the lengths of the corresponding sides are in the same proportion. And if the triangles have the lengths of the corresponding sides in the same proportion, then the corresponding angles are equal. In order to prove that two triangles are similar, rather than needing to prove that all corresponding angles are equal and all the lengths of corresponding sides are in proportion, there are a number of similarity criteria we can use.

The first criterion we could use is the angle–angle (AA) criterion.

Definition: Angle–Angle (AA) Similarity Criterion

If two pairs of corresponding angles in a pair of triangles are congruent, then the triangles are similar.

We can begin with the following pair of triangles.

Here, we are given that 𝑚𝐵=𝑚𝐸 and 𝑚𝐶=𝑚𝐹. We can show that with just two given pairs of angles equal, the third pair of angles, 𝑚𝐴 and 𝑚𝐷, must also be equal.

We can recall that the internal angles in a triangle sum to 180. Therefore, to work out the measure of angle 𝐴, we could calculate 𝑚𝐴+𝑚𝐵+𝑚𝐶=180𝑚𝐴=180(𝑚𝐵+𝑚𝐶).

In 𝐷𝐸𝐹, we could calculate the measure of 𝐷 as 𝑚𝐷+𝑚𝐸+𝑚𝐹=180𝑚𝐷=180(𝑚𝐸+𝑚𝐹).

Since we know that 𝑚𝐵=𝑚𝐸 and 𝑚𝐶=𝑚𝐹, we can say that 180(𝐵+𝐶)=180(𝐸+𝐹).

Therefore, 𝑚𝐴=𝑚𝐷.

Thus, when two pairs of corresponding angles in a triangle are congruent, the third pair of corresponding angles are also congruent and the triangles are similar.

The second similarity criterion involves the sides of the triangles.

Definition: Side-Side-Side (SSS) Similarity Criterion

If all three pairs of corresponding side lengths of two triangles are proportional, then the two triangles are similar.

We can apply this criterion in the following way. We can take the triangles 𝑃𝑄𝑅 and 𝐾𝐿𝑀.

If we can demonstrate that the sides in these triangles have a proportional relationship such that 𝑃𝑄𝐾𝐿=𝑄𝑅𝐿𝑀=𝑅𝑃𝑀𝐾, then 𝑃𝑄𝑅𝐾𝐿𝑀.

For example, we can consider the triangles below, 𝐴𝐵𝐶 and 𝐷𝐸𝐹.

We can write that 𝐴𝐵𝐷𝐸=𝐵𝐶𝐸𝐹=𝐴𝐶𝐷𝐹, since 63=105=94.5.

The proportions of corresponding sides are all equivalent to 21=2. Hence, 𝐴𝐵𝐶𝐷𝐸𝐹.

Note that it would also be valid to write 𝐷𝐸𝐴𝐵=𝐸𝐹𝐵𝐶=𝐷𝐹𝐴𝐶.

We simply need to ensure that we keep all the sides of each triangle either as the numerators or as the denominators.

As there is an equivalent proportional relationship between all corresponding side lengths of the triangle, the triangles are similar.

We will now discuss the final similarity criterion.

Definition: Side-Angle-Side (SAS) Similarity Criterion

If the lengths of two sides in one triangle are proportional to the lengths of two sides in another triangle and the included angles in both are congruent, then the two triangles are similar.

We can illustrate this criterion with the following figure.

Here, we have two pairs of corresponding side lengths in the same proportion, 𝐴𝐵𝑋𝑌=𝐵𝐶𝑌𝑍, since 106=63.6.

Both of these proportions are equivalent to 53. And, in both triangles, the included angles between these sides are congruent. Thus, 𝐴𝐵𝐶𝑋𝑌𝑍.

To fulfill the criterion for the SAS rule, we just need two pairs of sides in proportion, but the pair of angles have to be the angles between these sides in each triangle. Note that this rule is different from the SAS congruency criterion, where we must demonstrate that corresponding sides are equal for triangles to be congruent.

We will now see some examples of how we can apply these similarity criteria to prove that a pair of triangles are similar.

Example 1: Proving Whether Two Triangles are Similar

The figure shows a triangle 𝐴𝐷𝐸, where line segment 𝐵𝐶 is parallel to 𝐷𝐸.

  1. Which angle is equivalent to 𝐴𝐵𝐶? Why?
    1. 𝐴𝐷𝐸, because the angles are corresponding
    2. 𝐴𝐷𝐸, because the angles are alternate
    3. 𝐴𝐶𝐵, because the angles are alternate
    4. 𝐴𝐸𝐷, because the angles are corresponding
    5. 𝐴𝐶𝐵, because the angles are corresponding
  2. Which angle is equivalent to 𝐴𝐶𝐵? Why?
    1. 𝐴𝐵𝐶, because the angles are corresponding
    2. 𝐴𝐷𝐸, because the angles are corresponding
    3. 𝐴𝐷𝐸, because the angles are alternate
    4. 𝐴𝐸𝐷, because the angles are alternate
    5. 𝐴𝐸𝐷, because the angles are corresponding
  3. Hence, are triangles 𝐴𝐵𝐶 and 𝐴𝐷𝐸 similar? If yes, how?
    1. Yes, they are similar by the SSS criterion.
    2. Yes, they are similar by the SAS criterion.
    3. Yes, they are similar by the AA criterion.
    4. No, they are not similar.

Answer

Part 1

In the diagram, we observe that there is a pair of parallel line segments, 𝐵𝐶 and 𝐷𝐸. Line 𝐴𝐷 is a transversal of these; hence, the angle that is equal to 𝐴𝐵𝐶 is 𝐴𝐷𝐸.becausetheanglesarecorresponding

Part 2

To find the angle equivalent to 𝐴𝐶𝐵, we use the properties of the angles in parallel lines, along with the transversal 𝐴𝐸. Thus, the angle that is equivalent to 𝐴𝐶𝐵 is 𝐴𝐸𝐷.becausetheanglesarecorresponding

Part 3

We have now demonstrated that there are two pairs of corresponding angles equal in triangles 𝐴𝐵𝐶 and 𝐴𝐷𝐸: 𝑚𝐴𝐵𝐶=𝑚𝐴𝐷𝐸,𝑚𝐴𝐶𝐵=𝑚𝐴𝐸𝐷.

The AA similarity criterion states that if two pairs of corresponding angles in a pair of triangles are congruent, then the triangles are similar. Therefore, we can give the answer that triangles 𝐴𝐵𝐶 and 𝐴𝐷𝐸 are similar by the AA criterion.

We will now look at another example.

Example 2: Proving Whether Two Triangles are Similar

The figure shows two triangles 𝐴𝐵𝐶 and 𝐷𝐸𝐶, where line segment 𝐴𝐵 is parallel to 𝐷𝐸.

  1. Which angle is equivalent to 𝐴𝐵𝐶? Why?
    1. 𝐶𝐷𝐸, because the angles are corresponding
    2. 𝐶𝐷𝐸, because the angles are alternate
    3. 𝐶𝐸𝐷, because the angles are corresponding
    4. 𝐶𝐸𝐷, because the angles are alternate
    5. 𝐷𝐶𝐸, because the angles are vertically opposite
  2. Which angle is equivalent to 𝐵𝐴𝐶? Why?
    1. 𝐶𝐷𝐸, because the angles are alternate
    2. 𝐶𝐷𝐸, because the angles are corresponding
    3. 𝐶𝐸𝐷, because the angles are corresponding
    4. 𝐶𝐸𝐷, because the angles are alternate
    5. 𝐷𝐶𝐸, because the angles are vertically opposite
  3. Hence, are triangles 𝐴𝐵𝐶 and 𝐷𝐸𝐶 similar? If yes, how?
    1. Yes, they are similar by the SSS criterion.
    2. Yes, they are similar by the SAS criterion.
    3. Yes, they are similar by the AA criterion.
    4. No, they are not similar.

Answer

Part 1

In the diagram, the line segments 𝐴𝐵 and 𝐷𝐸 are marked as parallel. If we consider 𝐴𝐵𝐶, then, using the properties of the angles in parallel lines cut by a transversal, we can identify that the angle that is its equivalent is 𝐶𝐸𝐷.becausetheanglesarealternate

Part 2

Using the same properties, with the transversal 𝐴𝐷, the angle that is equivalent to 𝐵𝐴𝐶 is 𝐶𝐷𝐸.becausetheanglesarealternate

Part 3

We have shown that 𝑚𝐴𝐵𝐶=𝑚𝐶𝐸𝐷,𝑚𝐵𝐴𝐶=𝑚𝐶𝐷𝐸.

We can recall the AA similarity criterion, which states that if two pairs of corresponding angles in a pair of triangles are equal, then the triangles are similar. Hence, we can give the answer that triangles 𝐴𝐵𝐶 and 𝐷𝐸𝐶 are similar by the AA criterion.

We can generalize the methods used in the previous two questions in the following corollary of triangle similarity.

Definition: Corollary of Triangle Similarity

If the lengths of two sides in one triangle are proportional to the lengths of two sides in another triangle and the included angles in both are congruent, then the two triangles are similar.

In each of the figures above, we can state that if 𝐷𝐸𝐵𝐶 and intersects 𝐴𝐵 and 𝐴𝐶 at 𝐷 and 𝐸, respectively, then 𝐴𝐷𝐸𝐴𝐵𝐶.

In the following example, we will calculate the scale factor between a pair of similar triangles.

Example 3: Finding the Similarity Scale Factor

In the figure, 𝐶𝐵=5.7 and 𝐵𝐸=9.12. Since triangles 𝐶𝐵𝐴 and 𝐸𝐵𝐷 are similar, what is the scale factor?

Answer

We can begin by marking the given lengths onto the figure.

We are given that 𝐶𝐵𝐴 is similar to 𝐸𝐵𝐷. We can recall that similar triangles have corresponding pairs of angles that are congruent and corresponding side lengths that are in proportion.

We can use the ordering of the letters to help us identify that the lengths of 𝐶𝐵 and 𝐸𝐵 are corresponding. Thus, we can write the proportion of their lengths from 𝐶𝐵𝐴 to 𝐸𝐵𝐷 as 𝐸𝐵𝐶𝐵=9.125.7=85.

We can give the answer as a fraction or a decimal; thus, the scale factor from triangle 𝐶𝐵𝐴 to triangle 𝐸𝐵𝐷 is 1.6.

Note that a scale factor in the reverse direction, from 𝐸𝐵𝐷 to 𝐶𝐵𝐴, would be written as the proportion 𝐶𝐵𝐸𝐵=58.

We will now look at an example where we first need to prove that two triangles are similar and then use these properties to help us identify missing lengths.

Example 4: Calculating Unknowns Using Similarity and Finding the Perimeter of a Triangle

The figure shows triangle 𝐴𝐵𝐶.

  1. Work out the value of 𝑥.
  2. Work out the value of 𝑦.
  3. Work out the perimeter of 𝐴𝐵𝐶.

Answer

Within the larger triangle 𝐴𝐵𝐶, we observe that there is a smaller triangle, which we can label with points 𝐷 and 𝐸 to define 𝐴𝐷𝐸.

In order to find the missing lengths, we first determine if 𝐴𝐵𝐶 and 𝐴𝐷𝐸 are similar. Similar triangles have corresponding angles that are congruent and corresponding side lengths that are in proportion.

We note that in the figure, 𝐷𝐸 and 𝐵𝐶 are marked as parallel. This means that we can identify two pairs of corresponding angles using the transversals 𝐴𝐵 and 𝐴𝐶.

We can write that 𝐴𝐷𝐸=𝐴𝐵𝐶 and that 𝐴𝐸𝐷=𝐴𝐶𝐵.

The AA similarity criterion states that if two pairs of corresponding angles in a pair of triangles are congruent, then the triangles are similar. Alternatively, we could also have demonstrated that since angle 𝐴 is common to both triangles, then 𝐵𝐴𝐶=𝐷𝐴𝐸. Any two pairs out of these three pairs of angles would be sufficient to prove that 𝐴𝐷𝐸𝐴𝐵𝐶.

The proportion of the corresponding sides can be written as 𝐴𝐵𝐴𝐷=𝐵𝐶𝐷𝐸=𝐴𝐶𝐴𝐸.

We can now use this proportionality relationship to help us find the missing lengths, 𝑥 and 𝑦.

Part 1

The length, 𝑥, forms part of 𝐴𝐶, and the corresponding side in triangle 𝐴𝐷𝐸 is the side 𝐴𝐸.

When working with similar triangles, we usually have, or can calculate, the lengths of two corresponding sides. This allows us to find the proportion, or scale factor, between the given triangles. Here, we are given the lengths of 𝐷𝐸 and 𝐵𝐶. Hence, we can write that 𝐴𝐶𝐴𝐸=𝐵𝐶𝐷𝐸.

The length of 𝐴𝐶 can be written in terms of 𝑥 as (6+𝑥). We substitute in the values for the lengths to give 6+𝑥6=74.

Multiplying both sides by 6 and simplifying give 6+𝑥=424𝑥=4246𝑥=4.5.

Hence, we have found the value of 𝑥 as 4.5.

Part 2

The unknown length, 𝑦, is part of line segment 𝐴𝐵. The corresponding side in 𝐴𝐷𝐸 is 𝐴𝐷. We can use the proportionality relationship: 𝐴𝐵𝐴𝐷=𝐵𝐶𝐷𝐸.

We can represent length 𝐴𝐵 as (𝑦+5). Substituting in the values for the lengths and simplifying, we have 𝑦+55=74𝑦+5=354𝑦=3545=3.75.

Therefore, the value of 𝑦 is 3.75.

Part 3

The perimeter is the distance around the outside of a shape. Thus, to work out the perimeter of 𝐴𝐵𝐶, we have the following: perimeter=𝐴𝐵+𝐵𝐶+𝐶𝐴.

Substituting the lengths from the figure, we have perimeter=(5+𝑦)+7+(𝑥+6).

We calculated 𝑥=4.5 and 𝑦=3.75; therefore, we can simplify to give perimeter=5+3.75+7+4.5+6=26.25.

We can then give the answer that the perimeter of 𝐴𝐵𝐶 is 26.25.

In the following question, we will use the similarity of triangles to allow us to form and solve algebraic equations in order to find an unknown length.

Example 5: Forming and Solving an Equation Using Similarity to Find an Unknown

Triangles 𝐴𝐵𝐶 and 𝐴𝐷𝐸 are similar. Find 𝑥 to the nearest integer.

Answer

Within the larger triangle, 𝐴𝐵𝐶, in the figure, we observe that there is a smaller triangle, 𝐴𝐷𝐸. We cannot immediately calculate the lengths of 𝐸𝐷 and 𝐶𝐵; however, we can use the given information that these two triangles are similar.

In similar triangles, the proportion of sides is the same. Hence, we can write the proportion of corresponding sides as 𝐵𝐶𝐷𝐸=𝐴𝐶𝐴𝐸.

We can then substitute in the given lengths from the figure, being careful to note that 𝐴𝐶=5+2=7cm. This gives us 𝑥+24𝑥15=75.

Simplifying, we have 5(𝑥+2)=7(4𝑥15)5𝑥+10=28𝑥10510=23𝑥105115=23𝑥5=𝑥.

Therefore, we can give the answer that the value of 𝑥 is 5.

We will now see an example of how we can use the similarity of triangles to find measurements in a real-life situation. It is always helpful to first represent the given information in a sketch.

Example 6: Using the Similarity of Two Triangles to Find Indirect Measurements

A 1.97-metre-tall man stands 3.49 m away from a streetlight and casts a shadow that is 2.73 m long. How high is the lamp? Round your answer to the nearest tenth.

Answer

It might be useful to begin with a sketch of the information we have been given.

We can then model the situation using triangles, with a line joining the top of the light to the top of the shadow.

It is helpful to label the points. Here, we can define them as 𝑃, 𝑄, 𝑅, 𝐿, and 𝑀. We can assume that the man and the streetlight meet the horizontal ground at 90 and that the streetlight and the man are vertical and parallel.

We need to work out the height of the lamp, the length of 𝑃𝑄. If we had the length of 𝑃𝑅, we could apply the Pythagorean theorem. However, perhaps the best approach is to see if the triangle created with the light and the shadow, 𝑃𝑄𝑅, is similar to the smaller triangle created by the man and the shadow, 𝐿𝑀𝑅.

We recall that similar triangles have corresponding angles that are congruent and corresponding side lengths that are in proportion. One of the ways we can prove that two triangles are similar is by demonstrating that there are two pairs of corresponding angles that are congruent, that is, the AA criterion.

We can write that 𝑃𝑅𝑄=𝐿𝑅𝑀,𝑅,𝑃𝑄𝑅=𝐿𝑀𝑅,90.sinceiscommontobothtrianglesastheseareboth

As we have found two corresponding angles that are congruent, then we have proven that 𝑃𝑄𝑅𝐿𝑀𝑅.

In order to work out the length of 𝑃𝑄, we can use the corresponding side in 𝐿𝑀𝑅, side 𝐿𝑀. We know that the proportion between these sides will be the same as the proportion between the corresponding sides, 𝑄𝑅 and 𝑀𝑅, whose lengths we are given.

Therefore, 𝑃𝑄𝐿𝑀=𝑄𝑅𝑀𝑅.

We then substitute the length values, noting that 𝑄𝑅=3.49+2.73=6.22m. This gives us 𝑃𝑄1.97=6.222.73𝑃𝑄=6.222.73×1.97=4.4884.m

The height of the lamp was defined as 𝑃𝑄, and we round this value to the nearest tenth to give the answer for the height of the lamp, which is 4.5 m.

We can also note a particular instance of triangle similarity that involves an altitude of a right triangle.

Consider the following figure.

The largest triangle, right triangle 𝐴𝐵𝐷, is divided into two smaller triangles. Note that 𝐴𝐶 forms an altitude of this triangle, since it is the perpendicular drawn from the vertex of the triangle to the opposite side.

Note that as the sum of the angle measures on a straight line is 180, we can also state that 𝐴𝐶𝐷=90.

Let us consider the angles in these triangles, beginning with 𝐴𝐷𝐵. We can sketch the 3 triangles separately, with the right angle in the same position.

Since 𝐷 is a common angle between 𝐴𝐵𝐷 and 𝐶𝐴𝐷 and they both have a 90 angle, then by the AA similarity criterion, 𝐴𝐵𝐷𝐶𝐴𝐷.

Next, we observe that the angle at 𝐵 is common to both triangles 𝐴𝐵𝐷 and 𝐶𝐵𝐴.

Since these triangles also both have an angle of 90, then by the AA similarity criterion, 𝐴𝐵𝐷𝐶𝐵𝐴.

When any two triangles are each similar to a third triangle, then all three triangles are similar. Hence, 𝐴𝐵𝐷𝐶𝐴𝐷𝐶𝐵𝐴.

Hence, we can define a second corollary of triangle similarity below.

Definition: Corollary of Triangle Similarity

In any right triangle, the altitude to the hypotenuse separates the triangle into two triangles that are similar to each other and similar to the original triangle.

𝐴𝐵𝐷𝐶𝐴𝐷𝐶𝐵𝐴

We will now see how we can apply this corollary in the following example.

Example 7: Using the Similarity of Triangles Formed by the Altitude of a Right Triangle

The given figure shows a right triangle 𝐴𝐵𝐶, where 𝐶𝐷 is perpendicular to 𝐴𝐵.

  1. Using similarity, express 𝑎 in terms of 𝑐 and 𝑑.
  2. Using similarity, express 𝑏 in terms of 𝑐 and 𝑒.
  3. Express the sum of 𝑎 and 𝑏 in terms of 𝑐.

Answer

In the figure, we observe that 𝐶𝐷 is an altitude of the right triangle 𝐴𝐵𝐶. We can recall that in any right triangle, the altitude to the hypotenuse separates the triangle into two triangles that are similar to each other and similar to the original triangle.

Therefore, we can write that 𝐴𝐵𝐶𝐶𝐵𝐷𝐴𝐶𝐷.

Part 1

We can sketch the 3 triangles separately and in the same orientation with corresponding angles aligned.

As we need a relationship between the sides of lengths 𝑎, 𝑐, and 𝑑, we use 𝐴𝐵𝐶 and 𝐶𝐵𝐷 to form a similarity relationship.

We recognize that when triangles are similar, corresponding sides are in the same proportion. So, we have that 𝐶𝐵𝐷𝐵=𝐴𝐵𝐶𝐵𝑎𝑑=𝑐𝑎.

We then simplify this equation to give 𝑎=𝑐𝑑.

Thus, we have expressed 𝑎 in terms of 𝑐 and 𝑑.

Part 2

We use 𝐴𝐵𝐶 and 𝐴𝐶𝐷 to form the next similarity relationship, for the sides of lengths 𝑏, 𝑐, and 𝑒.

Hence, we have 𝐴𝐶𝐴𝐷=𝐴𝐵𝐴𝐶𝑏𝑒=𝑐𝑏𝑏=𝑐𝑒.

And so, we have 𝑏 expressed in terms of 𝑐 and 𝑒.

Part 3

We can write the sum of 𝑎 and 𝑏 by substituting the values 𝑎=𝑐𝑑 and 𝑏=𝑐𝑒 that we calculated in parts 1 and 2. This gives us 𝑎+𝑏=𝑐𝑑+𝑐𝑒=𝑐(𝑑+𝑒).

Using the figure, we can see that 𝑐=𝑑+𝑒; hence, 𝑎+𝑏=𝑐×𝑐=𝑐.

In fact, we may recognize this to be valid using the Pythagorean theorem. This states that for any right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides. Using similarity, we have demonstrated the validity of this theorem. We can express the sum of 𝑎 and 𝑏 in terms of 𝑐 as 𝑎+𝑏=𝑐.

We now summarize the key points.

Key Points

  • Two triangles are similar if corresponding angles are congruent and corresponding side lengths are in proportion.
  • We can prove that two triangles are similar using one of the following similarity criteria:
    • Angle–Angle (AA): If two pairs of corresponding angles in two triangles are congruent, then the triangles are similar.
    • Side-Side-Side (SSS): If all three pairs of corresponding side lengths of two triangles are proportional, then the two triangles are similar.
    • Side-Angle-Side (SAS): If two side lengths in one triangle are proportional to two side lengths in another triangle and the included angles in both are congruent, then the two triangles are similar.
  • In any right triangle, the altitude to the hypotenuse separates the triangle into two triangles that are similar to each other and similar to the original triangle.
  • We can model real-life situations involving similar triangles by drawing a sketch and using the proportionality of sides and the equivalency of angles to find unknown measurements.

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