Two figures are said to be similar if they have the exact same shape. It does not matter if the figures are different sizes as long as their shapes are the same. We can also say that two figures are similar if one is an exact copy of the other, drawn on a different scale. We come across the idea of scaling geometric objects all the time. When you construct a map, enlarge a photograph, or ask your computer to use a larger font size, you use scales. You simply create a new geometric object that has "the same shape" as the old one, but in a smaller or larger format. We will mainly touch on triangles, as similarity is mainly associated with this geometric figure in the context of competition, but of course similarity can be used with all geometric figures. We denote the area of the triangle $ ABC $ as $ [ABC] $.

The triangles below are similar and with the ratio $ 2: 1 $. If two triangles, $ \triangle ABC $ and $ \triangle DEF $, are similar, we write this as: $ \triangle ABC \sim \triangle DEF $. The triangles below are uniform and have the ratio $ 2: 1 $. The sides of the large triangle are thus $ 2 $ times larger than those in the small triangle.

Since we know this, we can calculate the length of $x $. Since we know that the ratio between the sides shall be $ 2: 1 $, we can conclude that $ \dfrac{| AC |}{| DF |} = \dfrac{x}{1} = \dfrac{| BC |}{ | EF |} = \dfrac{2}{1} $. This gives that $ x = 2 $. $ AC $ is called the corresponding side to the side $ DE $. Similarly, the side $ 5 $ corresponds to the side $ \dfrac {5} {2} $, and so on. The same goes for the angles. The angles that lie between two pairs of corresponding sides in each triangle are also called corresponding. For example, the angles $ \angle ACB $ and $ \angle DFE $ correspond.

If two triangles are similar, their corresponding angles are always equal. Furthermore, the ratio between corresponding side lengths is constant, i.e the ratio is always the same. For example, in the triangles above the ratio is $ \dfrac {| AB |} {| DE |} = \dfrac{| BC |}{| EF |} =2$. If we take the ratio of two other corresponding sides, the ratio will also be $2$. When two triangles are similar, and we know that a side in one triangle is twice as long in the corresponding side in the other triangle, then we know that all sides are twice as long. A third way we can define the same shape is as follows: two pairs of corresponding sides have the same ratio and the angle between them is the same in both triangles.

Thus, one usually talks about three different criteria or ways of identifying similarity, namely:

According to the Angle-Angle (AA) criterion, two triangles are similar if two angles in one triangle are equal to two angles in the other triangle. Since the sum of the angles in a triangle is $ 180^{\circ} $, we can always calculate the third angle if we know two angles. Therefore, it is sufficient that two angles are the same in the two triangles, because the third is then automatically also the same. In mathematical terms we can express this as:

If $ \triangle ABC $ and $ \triangle XYZ $ are two triangles where $ \angle {A} = \angle {X} $ and $ \angle {B} = \angle {Y} $, then $ \triangle ABC \sim \triangle DEF $.

Let us say that we have a triangle with the side lengths $4,5$ and $6$, and that we have another larger triangle with the side lengths $8,10$ and $12$. According to the Side-Side-Side (SSS) criterion, these two triangles are similar because the ratio between each pair of corresponding sides is the same. $ 8$ is twice as large as $ 4 $, $ 10$ is twice as large as $ 5 $ and $ 12 $ is twice as large as $ 6 $. Of course, the ratio does not have to be exactly $ 2: 1 $. The important thing is just that the ratio is equal between each pair of corresponding sides. When the relationship between two pairs of sides is the same, we say that the sides are proportional to each other. Therefore, two triangles are similar if all three sides of a triangle are proportional to the sides of the other triangle. In mathematical terms, we can express the SSS criterion as:

If $ \triangle ABC $ and $ \triangle XYZ $ are two triangles such that $ \dfrac {| AB |} {| XY |} = \dfrac {| AC |} {| XZ |} = \dfrac {| BC | } {| YZ |} $, then $ \triangle ABC \sim \triangle DEF $.

where $a>0$.

According to the Side-Angle-Side (SAS) criterion, two triangles are similar if two sides of one triangle are proportional to two sides of the other triangle and that the angle between these sides is equal in both triangles. Let's say we have two triangles, one with the side lengths $ 4 $ and $ 7 $ and the other with the side lengths $ 12 $ and $ 21 $, and with angle $ 40 ^ {\circ} $ between the sides. Since $ 12$ is three times as large as $ 4 $ and $ 21 $ is three times as large as $ 7 $, and the angles between the sides are equal, the triangles are similar. In mathematical terms we can express this as:

If $ \triangle ABC $ and $ \triangle XYZ $ are two triangles such that $ \angle {A} = \angle {X} $ and $ \dfrac {| AB |} {| XY |} = \dfrac {| AC |} {| XZ |} $, then $ \triangle ABC \sim \triangle DEF $.

This example illustrates Exercise 8. 

If we know that both triangles are right angles, it is enough that two pairs of sides have the same ratio for the triangles to be similar. In fact, it does not matter if the right angle is between the corresponding sides or not. The triangles must have the same shape anyway. If the right angle is between the two corresponding, we are back in the SVS criterion. There is only one triangle with two specific side lengths where the angle between these lengths is 90 degrees. If, on the other hand, the right angle is not between the two corresponding, we are back in the SSV criterion. But since the side can now not oscillate (we have a right angle), it will only be one possible triangle. The right angle must therefore not necessarily be between the two pairs of corresponding sides. The triangles will be similar anyway. We can therefore write an extra criterion for the special case when the triangles have right angles:

If $ \triangle ABC $ and $ \triangle XYZ $ are two right-angled triangles such that $ \dfrac {| AB |} {| XY |} = \dfrac {| AC |} {| XZ |} $, then $ \triangle ABC \sim \triangle DEF $.

Case 1)

where $a>0$.

Case 2)

where $a>0$.

There are different ways to prove similarity and if we have proven one criterion, we automatically know the others. Since all criteria are really the same, it means that we can choose what suits the task best. The AA-criterion is the most commonly used criterion. It becomes even more interesting when you start to find your own criteria for identifying similarity and congruence. For example, one can draw conclusions based on areas and perimeters, which can be useful tools in competition contexts.

The most common criterion used is AA, as I previously mentioned. This means that the concept of similar triangles is basically about angle chasing (see previous lesson). It turns out that parallel lines often create similar triangles. This is because if we have two lines that intersect and then two parallel lines that intersect them again, the angle will be the same. There are two main cases of how similar triangles can occur with parallel lines, namely:

$\bullet$ When two lines intersect and the point of intersection lies between two parallel lines.

$\bullet$ When two lines intersect and the point of intersection is outside two parallel lines.

These are illustrated below.

Case 1)

Case 2)

The two red lines are parallel and intersect the black lines at $ B, C, F, $ and $ E $. The triangle $ AEF $ in the first figure will then be similar to the triangle $ ABC $. This is because $ \angle AEF = \angle ABC $ and $ \angle BCA = \angle EFA $. By the AA criterion, they are thus similar. In the second figure, $ \triangle ABC \sim \triangle AEF $ due to alternate angles, ie the AA-criterion.

Below are some small but important tools that I think everyone should learn. I also encourage you to derive these tools before you begin the exercises, to gain a deeper understanding of the subject; it is important to get to know ones tools.

For a given triangle $ ABC $, let $ M $ and $ N $ be the midpoints of $ AC $ and $ AB $, respectively. Then the segment $ MN $ is parallel to $ BC $ and its length is half the length of $ BC $.

For a given triangle $ ABC $, let $ E $ and $ F $ be points on the sides $ AC $ and $ AB $, respectively, such that $ EF \parallel BC $. Then $ \triangle ABC \sim \triangle AEF $.

For a given triangle $ ABC $, let $ E $ and $ F $ be points on the sides $ AC $ and $ AB $, respectively, such that $ EF \parallel BC $. Then $ \dfrac {| AE |} {| EC |} = \dfrac {| AF |} {| FB |} $.