If you need help with hash marks and bands and terms like corresponding parts or the included angle, I have explained them in the footnotes.

In this tutorial, we will learn all about similar triangles – what similarity of triangles means, how to identify similar triangles, and also how the concept is useful.

Two triangles are said to be similar if they have the exact same shape. They may or may not have the same size.

One way to think of similarity is – if one triangle can be turned into another by scaling it up or down (zooming in or out) and adjusting its orientation.

For example, if you flip the green triangle sideways and scale it up by 50% (1.5x), it will overlap perfectly with the yellow one.

To determine whether two triangles are similar, we have three sets of conditions to check against. If we can prove that the two triangles fulfill any one set of conditions, we can be sure they are similar.

Here are the three criteria.

By the AA condition, two triangles are similar if two angles in one triangle are the same as two angles in the other triangle.

By the SSS condition, two triangles are similar if the three sides in one triangle are proportional to the three sides in the other triangle.

By the SAS condition, two triangles are similar if two sides in one triangle are proportional to two sides in the other triangle and the included angles (between the two sides in each triangle) are equal.

The symbol for similarity is ~. However, there’s more to it than just the symbol.

When we write the names of the two triangles, corresponding vertices must be in the same order/position.

For example, in the figure below vertices A, B, and C correspond to Q, P, and R respectively (corresponding angles are equal).

Remember, corresponding vertices have equal angles and sides opposite to corresponding vertices are proportional.

So the correct notation for their similarity would be △ ABC ~ △ QPR.

Here are the three most commonly used properties of similar triangles. We’ll use the triangles below as examples.

Property 1. Corresponding angles are equal.

$\begin{align*} m(\angle A) = m(\angle Y) \\[1em] m(\angle B) = m(\angle Z) \\[1em] m(\angle C) = m(\angle X) \end{align*}$

Property 2. The three pairs of corresponding sides are in proportion.

$\frac{AB}{YZ} = \frac{BC}{ZX} = \frac{AC}{YX}$

Property 3. The ratio of areas is equal to the square of the ratio of corresponding sides.

$\frac{\text{Area }(\triangle ABC)}{\text{Area }(\triangle YZX)} = \left ( \frac{AB}{YZ} \right )^2$

Alright, now that we know the criteria for similarity and the properties of similar triangles, it’s time to solve a couple of examples.

Example

Using the information provided in the figure below, prove that the two triangles are similar. And hence, find the length of PR.

Solution

To prove two triangles are similar, we need to see which of the three criteria would be the simplest or most convenient to apply.

Here, in △ ABC and △ PQR,

$\begin{align*} \angle A &\cong \angle Y \\[1em] \angle B &\cong \angle Z \end{align*}$

So, △ ABC ~ △ RQP by AA similarity. And we have proved the two triangles are similar.

Now, since corresponding sides of similar triangles are in proportion, we have –

$\frac{RQ}{AB} = \frac{QP}{BC} = \frac{RP}{AC}$

Substituting the values from the question,

$\frac{RQ}{15} = \frac{9}{12} = \frac{RP}{20}$

Let’s rewrite this as two equations from which we can find the unknown sides – RQ and RP.

$\begin{align*} \frac{RQ}{15} &= \frac{9}{12} \\[1.3em] RQ &= 11.25 \end{align*}$

$\begin{align*} \frac{9}{12} &= \frac{RP}{20} \\[1.3em] RP &= 15 \end{align*}$

That's it.

Example

Prove that triangles ABC and CDE are similar. Also, if the area of △ ABC is 69.28 in2, what is the area of △ CDE?

Solution

In this question, we know two sides of each triangle. So that gives us a clue into what we need to do.

Let’s find the ratio between the smaller sides (one from each triangle) and between the two larger sides.

$\begin{align*} \frac{CD}{BC} &= \frac{10}{20} = 2 \hspace{0.25cm} \rule[0.1cm]{1cm}{0.1em} \hspace{0.15cm} (1) \\[1.3em] \frac{CE}{AC} &= \frac{16}{32} = 2 \hspace{0.25cm} \rule[0.1cm]{1cm}{0.1em} \hspace{0.15cm} (2) \end{align*}$

As expected, the two ratios are equal.

$\frac{CD}{BC} = \frac{CE}{AC}$

This means the two pairs of sides are in proportion. Also, the included angles are equal (they are vertically opposite angles).

$\angle ECD \cong \angle BCA$

So, △ ABC ~ △ EDC by SAS similarity.

Now, we know that the ratio of areas of two similar triangles is equal to the square of the ratios of corresponding sides. So,

$\frac{\text{Area }(\triangle EDC)}{\text{Area }(\triangle ABC)} = \left ( \frac{CD}{BC} \right )^2$

Substituting the values from the question, we have –

$\begin{align*} \frac{\text{Area }(\triangle EDC)}{69.28} &= 2^2 \\[1.3em] \text{Area }(\triangle EDC) &= 4 \times 69.28 \\[1.3em] &= 277.12 \end{align*}$

So the area of $\hspace{0.2em} \triangle CDE \, \hspace{0.2em}$ is $\hspace{0.2em} \,277.12 \text{ in}^2 \hspace{0.2em}$.

Similarity and congruency are two closely related concepts. So let’s see what’s common and what’s different between them.

The idea is simple. Similarity requires two triangles (or any geometric figures) to have exactly the same shape. They may or may not have the same size. Congruency, on the other hand, requires them to have exactly the same shape and size.

So if two triangles are congruent, they must be similar too. But the converse is not true.

And that brings us to the end of this tutorial on similar triangles. Until next time.

Corresponding parts (sides or angles) refer to parts in two triangles that have the same relative position - sort of matching parts.

Here's an example.

In the figure below, the second triangle is a scaled-up (1.5x) and flipped version of the first triangle. Also, if you imagine the transformation, vertex C in the first triangle turned into P in the second, and side BC turned into PR.

So which side in the second triangle corresponds to AB in the first? And which angle corresponds to B?

Well, AB is opposite C. And C and P are corresponding angles. So QR – the side opposite P – must correspond to AB.

Now, side BC corresponds to side PR. And C (one angle on BC) corresponds to P (one angle on PR). So B (the other angle on BC) and R (the other angle on PQ) must be corresponding parts.

In the figure below, the corresponding sides/angles are in the same color.

For any two sides in a triangle (or a polygon), the included angle is the one formed between those sides – at their intersection. For example, here B is the included angle between AB and BC.

Similarly, the included side is the side that’s common to two angles. In the figure below, AC is the included side for angles A and C.

When comparing two triangles or even two sides or angles within a triangle, hash marks and/or bands are used to show which sides or angles are equal. For example –

We use cookies to provide and improve our services. By using the site you agree to our use of cookies. Learn more