The portal has been deactivated. Please contact your portal admin.

Lesson Explainer: Triangle Similarity Criteria and Their Applications Mathematics

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, 𝐡 and 𝐸, and 𝐢 and 𝐹, are of equal measures. The lengths of the corresponding sides, 𝐴𝐡 and 𝐷𝐸, 𝐴𝐢 and 𝐷𝐹, and 𝐡𝐢 and 𝐸𝐹, are in the same proportion. In this example, we can also say that the scale factor from △𝐴𝐡𝐢 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 π‘˜ would produce the following triangle 𝐹′𝐺′𝐻′.

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 𝐹′𝐺′𝐻′, and we can write this as β–³πΉπΊπ»βˆΌβ–³πΉβ€²πΊβ€²π»β€².

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 π‘šβˆ π΅=π‘šβˆ πΈ and π‘šβˆ πΆ=π‘šβˆ πΉ. We can show that with just two given pairs of angles equal, the third pair of angles, π‘šβˆ π΄ and π‘šβˆ π·, 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 π‘šβˆ π΄+π‘šβˆ π΅+π‘šβˆ πΆ=180π‘šβˆ π΄=180βˆ’(π‘šβˆ π΅+π‘šβˆ πΆ).∘∘

In △𝐷𝐸𝐹, we could calculate the measure of ∠𝐷 as π‘šβˆ π·+π‘šβˆ πΈ+π‘šβˆ πΉ=180π‘šβˆ π·=180βˆ’(π‘šβˆ πΈ+π‘šβˆ πΉ).∘∘

Since we know that π‘šβˆ π΅=π‘šβˆ πΈ and π‘šβˆ πΆ=π‘šβˆ πΉ, we can say that 180βˆ’(∠𝐡+∠𝐢)=180βˆ’(∠𝐸+∠𝐹).∘∘

Therefore, π‘šβˆ π΄=π‘šβˆ π·.

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 β–³π‘ƒπ‘„π‘…βˆΌβ–³πΎπΏπ‘€.

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

We can write that 𝐴𝐡𝐷𝐸=𝐡𝐢𝐸𝐹=𝐴𝐢𝐷𝐹, since 63=105=94.5.

The proportions of corresponding sides are all equivalent to 21=2. Hence, β–³π΄π΅πΆβˆΌβ–³π·πΈπΉ.

Note that it would also be valid to write 𝐷𝐸𝐴𝐡=𝐸𝐹𝐡𝐢=𝐷𝐹𝐴𝐢.

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, π΄π΅π‘‹π‘Œ=π΅πΆπ‘Œπ‘, 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, β–³π΄π΅πΆβˆΌβ–³π‘‹π‘Œπ‘.

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 𝐡𝐢 is parallel to 𝐷𝐸.

  1. Which angle is equivalent to ∠𝐴𝐡𝐢? Why?
    1. ∠𝐴𝐷𝐸, because the angles are corresponding
    2. ∠𝐴𝐷𝐸, because the angles are alternate
    3. ∠𝐴𝐢𝐡, because the angles are alternate
    4. ∠𝐴𝐸𝐷, because the angles are corresponding
    5. ∠𝐴𝐢𝐡, because the angles are corresponding
  2. Which angle is equivalent to ∠𝐴𝐢𝐡? Why?
    1. ∠𝐴𝐡𝐢, 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 𝐴𝐡𝐢 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, 𝐡𝐢 and 𝐷𝐸. Line ⃖⃗𝐴𝐷 is a transversal of these; hence, the angle that is equal to ∠𝐴𝐡𝐢 is ∠𝐴𝐷𝐸.becausetheanglesarecorresponding

Part 2

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

Part 3

We have now demonstrated that there are two pairs of corresponding angles equal in triangles 𝐴𝐡𝐢 and 𝐴𝐷𝐸: π‘šβˆ π΄π΅πΆ=π‘šβˆ π΄π·πΈ,π‘šβˆ π΄πΆπ΅=π‘šβˆ π΄πΈπ·.

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 𝐴𝐡𝐢 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 𝐴𝐡𝐢 and 𝐷𝐸𝐢, where line segment 𝐴𝐡 is parallel to 𝐷𝐸.

  1. Which angle is equivalent to ∠𝐴𝐡𝐢? Why?
    1. ∠𝐢𝐷𝐸, because the angles are corresponding
    2. ∠𝐢𝐷𝐸, because the angles are alternate
    3. ∠𝐢𝐸𝐷, because the angles are corresponding
    4. ∠𝐢𝐸𝐷, because the angles are alternate
    5. ∠𝐷𝐢𝐸, because the angles are vertically opposite
  2. Which angle is equivalent to ∠𝐡𝐴𝐢? Why?
    1. ∠𝐢𝐷𝐸, because the angles are alternate
    2. ∠𝐢𝐷𝐸, because the angles are corresponding
    3. ∠𝐢𝐸𝐷, because the angles are corresponding
    4. ∠𝐢𝐸𝐷, because the angles are alternate
    5. ∠𝐷𝐢𝐸, because the angles are vertically opposite
  3. Hence, are triangles 𝐴𝐡𝐢 and 𝐷𝐸𝐢 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 𝐴𝐡 and 𝐷𝐸 are marked as parallel. If we consider ∠𝐴𝐡𝐢, 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 ∠𝐢𝐸𝐷.becausetheanglesarealternate

Part 2

Using the same properties, with the transversal 𝐴𝐷, the angle that is equivalent to ∠𝐡𝐴𝐢 is ∠𝐢𝐷𝐸.becausetheanglesarealternate

Part 3

We have shown that π‘šβˆ π΄π΅πΆ=π‘šβˆ πΆπΈπ·,π‘šβˆ π΅π΄πΆ=π‘šβˆ πΆπ·πΈ.

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 𝐴𝐡𝐢 and 𝐷𝐸𝐢 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 ⃖⃗𝐷𝐸⫽𝐡𝐢 and intersects ⃖⃗𝐴𝐡 and ⃖⃗𝐴𝐢 at 𝐷 and 𝐸, respectively, then β–³π΄π·πΈβˆΌβ–³π΄π΅πΆ.

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, 𝐢𝐡=5.7 and 𝐡𝐸=9.12. Since triangles 𝐢𝐡𝐴 and 𝐸𝐡𝐷 are similar, what is the scale factor?

Answer

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

We are given that △𝐢𝐡𝐴 is similar to △𝐸𝐡𝐷. 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 𝐢𝐡 and 𝐸𝐡 are corresponding. Thus, we can write the proportion of their lengths from △𝐢𝐡𝐴 to △𝐸𝐡𝐷 as 𝐸𝐡𝐢𝐡=9.125.7=85.

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

Note that a scale factor in the reverse direction, from △𝐸𝐡𝐷 to △𝐢𝐡𝐴, would be written as the proportion 𝐢𝐡𝐸𝐡=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 𝐴𝐡𝐢.

  1. Work out the value of π‘₯.
  2. Work out the value of 𝑦.
  3. Work out the perimeter of △𝐴𝐡𝐢.

Answer

Within the larger triangle 𝐴𝐡𝐢, 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 △𝐴𝐡𝐢 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 𝐡𝐢 are marked as parallel. This means that we can identify two pairs of corresponding angles using the transversals ⃖⃗𝐴𝐡 and ⃖⃗𝐴𝐢.

We can write that ∠𝐴𝐷𝐸=∠𝐴𝐡𝐢 and that ∠𝐴𝐸𝐷=∠𝐴𝐢𝐡.

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 ∠𝐡𝐴𝐢=∠𝐷𝐴𝐸. Any two pairs out of these three pairs of angles would be sufficient to prove that β–³π΄π·πΈβˆΌβ–³π΄π΅πΆ.

The proportion of the corresponding sides can be written as 𝐴𝐡𝐴𝐷=𝐡𝐢𝐷𝐸=𝐴𝐢𝐴𝐸.

We can now use this proportionality relationship to help us find the missing lengths, π‘₯ and 𝑦.

Part 1

The length, π‘₯, forms part of 𝐴𝐢, 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 𝐡𝐢. Hence, we can write that 𝐴𝐢𝐴𝐸=𝐡𝐢𝐷𝐸.

The length of 𝐴𝐢 can be written in terms of π‘₯ as (6+π‘₯). We substitute in the values for the lengths to give 6+π‘₯6=74.

Multiplying both sides by 6 and simplifying give 6+π‘₯=424π‘₯=424βˆ’6π‘₯=4.5.

Hence, we have found the value of π‘₯ as 4.5.

Part 2

The unknown length, 𝑦, is part of line segment 𝐴𝐡. The corresponding side in △𝐴𝐷𝐸 is 𝐴𝐷. We can use the proportionality relationship: 𝐴𝐡𝐴𝐷=𝐡𝐢𝐷𝐸.

We can represent length 𝐴𝐡 as (𝑦+5). Substituting in the values for the lengths and simplifying, we have 𝑦+55=74𝑦+5=354𝑦=354βˆ’5=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 △𝐴𝐡𝐢, we have the following: perimeter=𝐴𝐡+𝐡𝐢+𝐢𝐴.

Substituting the lengths from the figure, we have perimeter=(5+𝑦)+7+(π‘₯+6).

We calculated π‘₯=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 △𝐴𝐡𝐢 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 𝐴𝐡𝐢 and 𝐴𝐷𝐸 are similar. Find π‘₯ to the nearest integer.

Answer

Within the larger triangle, 𝐴𝐡𝐢, in the figure, we observe that there is a smaller triangle, 𝐴𝐷𝐸. We cannot immediately calculate the lengths of 𝐸𝐷 and 𝐢𝐡; 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 𝐡𝐢𝐷𝐸=𝐴𝐢𝐴𝐸.

We can then substitute in the given lengths from the figure, being careful to note that 𝐴𝐢=5+2=7cm. This gives us π‘₯+24π‘₯βˆ’15=75.

Simplifying, we have 5(π‘₯+2)=7(4π‘₯βˆ’15)5π‘₯+10=28π‘₯βˆ’10510=23π‘₯βˆ’105115=23π‘₯5=π‘₯.

Therefore, we can give the answer that the value of π‘₯ 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 βˆ π‘ƒπ‘…π‘„=βˆ πΏπ‘…π‘€,βˆ π‘…,βˆ π‘ƒπ‘„π‘…=βˆ πΏπ‘€π‘…,90.sinceiscommontobothtrianglesastheseareboth∘

As we have found two corresponding angles that are congruent, then we have proven that β–³π‘ƒπ‘„π‘…βˆΌβ–³πΏπ‘€π‘….

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=6.222.73Γ—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 𝐴𝐡𝐷, is divided into two smaller triangles. Note that 𝐴𝐢 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 ∠𝐴𝐢𝐷=90∘.

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

Since 𝐷 is a common angle between △𝐴𝐡𝐷 and △𝐢𝐴𝐷 and they both have a 90∘ angle, then by the AA similarity criterion, β–³π΄π΅π·βˆΌβ–³πΆπ΄π·.

Next, we observe that the angle at 𝐡 is common to both triangles 𝐴𝐡𝐷 and 𝐢𝐡𝐴.

Since these triangles also both have an angle of 90∘, then by the AA similarity criterion, β–³π΄π΅π·βˆΌβ–³πΆπ΅π΄.

When any two triangles are each similar to a third triangle, then all three triangles are similar. Hence, β–³π΄π΅π·βˆΌβ–³πΆπ΄π·βˆΌβ–³πΆπ΅π΄.

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.

β–³π΄π΅π·βˆΌβ–³πΆπ΄π·βˆΌβ–³πΆπ΅π΄

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 𝐴𝐡𝐢, where 𝐢𝐷 is perpendicular to 𝐴𝐡.

  1. Using similarity, express π‘ŽοŠ¨ in terms of 𝑐 and 𝑑.
  2. Using similarity, express π‘οŠ¨ in terms of 𝑐 and 𝑒.
  3. Express the sum of π‘ŽοŠ¨ and π‘οŠ¨ in terms of 𝑐.

Answer

In the figure, we observe that 𝐢𝐷 is an altitude of the right triangle 𝐴𝐡𝐢. 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 β–³π΄π΅πΆβˆΌβ–³πΆπ΅π·βˆΌβ–³π΄πΆπ·.

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 π‘Ž, 𝑐, and 𝑑, we use △𝐴𝐡𝐢 and △𝐢𝐡𝐷 to form a similarity relationship.

We recognize that when triangles are similar, corresponding sides are in the same proportion. So, we have that 𝐢𝐡𝐷𝐡=π΄π΅πΆπ΅π‘Žπ‘‘=π‘π‘Ž.

We then simplify this equation to give π‘Ž=𝑐𝑑.

Thus, we have expressed π‘ŽοŠ¨ in terms of 𝑐 and 𝑑.

Part 2

We use △𝐴𝐡𝐢 and △𝐴𝐢𝐷 to form the next similarity relationship, for the sides of lengths 𝑏, 𝑐, and 𝑒.

Hence, we have 𝐴𝐢𝐴𝐷=𝐴𝐡𝐴𝐢𝑏𝑒=𝑐𝑏𝑏=𝑐𝑒.

And so, we have π‘οŠ¨ expressed in terms of 𝑐 and 𝑒.

Part 3

We can write the sum of π‘ŽοŠ¨ and π‘οŠ¨ by substituting the values π‘Ž=π‘π‘‘οŠ¨ and 𝑏=π‘π‘’οŠ¨ that we calculated in parts 1 and 2. This gives us π‘Ž+𝑏=𝑐𝑑+𝑐𝑒=𝑐(𝑑+𝑒).

Using the figure, we can see that 𝑐=𝑑+𝑒; hence, π‘Ž+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 π‘ŽοŠ¨ and π‘οŠ¨ in terms of 𝑐 as π‘Ž+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.

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy.