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## What are Proportionality Theorems in Geometry?

Proportionality theorems show relationships between shapes in the form of ratios. They show how different ratios of a figure or a quantity are equal. The proportionality theorems are mostly used in triangles. Let's look at the fundamental concept of the proportionality theorem using the triangle figures below.

The triangles above will be called **similar** triangles if their angles are congruent and if their corresponding sides are proportional. So, the proportionality formula for similar triangles is below.

$\frac{AB}{KL}=\frac{AC}{KM}=\frac{BC}{LM}$

## What is the Basic Proportionality Theorem?

The Basic Proportionality Theorem focuses on showing the relationship between the length of the sides of a triangle.

The proportionality theorem states that if a line is drawn parallel to one side of a triangle to intersect the other two sides at distinct points, then the other two sides are divided in the same ratio.

The figure below gives a visual representation of the theorem.

In the $\u2206ABC$ above, $\overline{DE}$ is parallel to $\overline{BC}$. According to the Basic Proportionality Theorem, the ratio of $\overline{AD}$ to $DB$ is equal to the ratio of $\overline{AE}$ to $EC$:

$\frac{AD}{DB}=\frac{AE}{EC}$The ratio above is considered the basic proportionality formula.

We can prove this theorem and find out how to get the formula. Let's see how.

From the theorem, we know that $DB$ and $EC$ are in the same ratio and we want to prove that they are equal. We will first form triangles that have $DB$ and $EC$ as their side lengths. To get these triangles, we will draw a segment joining $B$ to $E$ and another segment joining $C$ to $D$ as shown below.

We have now formed two new triangles$(\u2206DEBand\u2206DEC)$.

The next thing is to find a relationship between the new triangles. In particular, let's look at the area. $\u2206DEBand\u2206DEC$ have the same base $\overline{DE}$ and the same height because the third vertex of the triangle is between the same parallel. Therefore, the area of both triangles must be equal:

$Are{a}_{(\u2206DEB)}=Are{a}_{(\u2206EDC)}$Now, consider $\u2206AED$. Let's take $AD$ as the base and the height as the perpendicular distance from the line $AD$ to the opposite vertex $E$. See how it looks like in the figure below.

The area of this triangle is

$Are{a}_{(\u2206AED)}=\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\times \overline{AD}\times \overline{EP}$We also need the area of $\u2206DEB$ which will be:

$Are{a}_{(\u2206DEB)}=\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\times \overline{DB}\times \overline{EP}$Now, we can take the ratio of the area of $\u2206DEB$ to the area of $\u2206AED$ and compare it with the ratio of the area of $\u2206ECD$ to the area of $\u2206AED$. Therefore, the ratio of the areas is:

$\frac{ar(\u2206AED)}{ar(\u2206DEB)}=\frac{\overline{)\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\times AD\times \overline{)EP}}{\overline{)\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\times DB\times \overline{)EP}}=\frac{AD}{DB}$

As you can see, we've got the first part of the formula. To get the other, we will repeat everything we just did but with$\u2206EDC$.

Unlike before, instead of using $AD$ as the base of $\u2206AED$, we will use $AE$ as the base and the height will be the perpendicular distance opposite the vertex $D$. See how it looks like in the figure below.

The area of $\u2206AED$ according to the image above is

$Are{a}_{(\u2206AED)}=\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\times \overline{AE}\times \overline{DQ}$

Now let's consider the area of $\u2206EDC$. We will take EC as the base and $DQ$ as the height. The area is as follows.

$Are{a}_{(\u2206EDC)}=\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\times \overline{EC}\times \overline{DQ}$

We will now get the ratio of both areas to be:

$\frac{Are{a}_{(\u2206AED)}}{Are{a}_{(\u2206EDC)}}=\frac{\overline{)\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\times AE\times \overline{)DQ}}{\overline{)\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\times EC\times \overline{)DQ}}=\frac{AE}{EC}$

So, you can see that we've gotten the other part of the formula. But how do we show that both parts are equal? Let's equate both ratios and see.

$\frac{Are{a}_{(\u2206AED)}}{Are{a}_{(\u2206DEB)}}=\frac{Are{a}_{(\u2206AED)}}{Are{a}_{(\u2206EDC)}}$

Both numerators are the same so they are equal. Recall that at the beginning of the proof, we saw that

$Are{a}_{(\u2206DEB)}=Are{a}_{(\u2206EDC)}$Therefore,

$\frac{AD}{DB}=\frac{AE}{EC}$### The Triangle Proportionality Theorem and Fundamental Theorem of Proportionality

The T**riangle Proportionality Theorem of Fundamental Theorem of Proportionality are just other names for the Basic Proportionality Theorem. You may see this theorem referred to as any of these titles!**

## Proportionality Theorem Examples

Let's see the application of the proportionality theorem with some examples.

Consider a $\u2206ABC$ where DE is parallel to $BC$. $AD=1.5cm,DB=3cm,AE=1$. Find $EC$.

Remember the formula

$\frac{AD}{DB}=\frac{AE}{EC}$All we have to do is substitute the values.

$\frac{1.5}{3}=\frac{1}{EC}\phantom{\rule{0ex}{0ex}}1.5\times EC=3\times 1\phantom{\rule{0ex}{0ex}}1.5EC=3\phantom{\rule{0ex}{0ex}}EC=\frac{3}{1.5}\phantom{\rule{0ex}{0ex}}EC=2cm$

Let's take a look at another example.

Consider $\u2206EFG$ where $HL$ and $EF$ are parallel to each other. $EH=9cm,HG=21,FL=6cm$. Find $LG$

According to the proportionality theorem,

$\frac{EH}{HG}=\frac{FL}{LG}$Subbing in the known values leaves us with

$\begin{array}{rcl}\frac{9}{21}& =& \frac{6}{LG}\\ 9\times LG& =& 6\times 21\\ 9LG& =& 126\\ LG& =& \frac{126}{9}\\ LG& =& 14cm\end{array}$

Aside from showing the relationship between the length of sides of triangles, in real life, the proportionality theorem can be used in construction.

## The converse of the Basic Proportionality Theorem

The converse of the basic proportionality theorem is the reverse of the basic proportionality theorem. The theorem states that if a line is drawn to intersect two sides of a triangle at different points such that it cuts the two sides in the same ratio, then the line is parallel to the third side.

In the basic proportionality theorem, we saw that $DE$ and $BC$ are parallel and now we want to prove that $DE$ and $BC$ are indeed parallel. We would do this using the basic proportionality theorem which is

$\frac{AD}{DB}=\frac{AE}{EC}$This proof is proof by contradiction meaning that we will assume that our desired result is wrong. We will assume that $DE$ is not parallel to $BC$ ($(DE\nparallel BC)$. If this is the case, then there must be another point on line $AC$ such that a segment drawn from point $D$ to that point is parallel to $BC$. See the figure below for clarity.

Now that we have a line segment $AF$ that is parallel to $BC$, we can now use the basic proportionality theorem which is

$\frac{AD}{DB}=\frac{AF}{FC}$

If you consider the basic proportionality theorem, you will have:

$\frac{AD}{DB}=\frac{AF}{FC}=\frac{AE}{EC}$

We have now derived that $DF$ is parallel to $BC$ and we want to show that $DE$ is parallel to $BC$. This means that what we really want to do is show that $DF$ and $DE$ are the same segments. So, if we consider the above equation, you will see that the first ratio is not really needed. So we are left with

$\frac{AF}{FC}=\frac{AE}{EC}$

We are now saying that $DF$ and $DE$ are the same segments which means that point $F$ and point $E$ are the same. If this is our conclusion, then the segment $AF$ and $AE$ are the same but we haven't really proven that yet.

From the figure, we can say that the segment $AC$ is equal to the sum of the segment $AE$ and $EC$.

$AC=AE+EC$

Let's go back to one of our equations.

$\frac{AF}{FC}=\frac{AE}{EC}$We will now add **1 **(one)** **to both sides of the equation and bring them into the fractions by giving them a common denominator.

$\begin{array}{rcl}\frac{AF}{FC}+1& =& \frac{AE}{EC}+1\\ & & \\ \frac{AF}{FC}+\frac{FC}{FC}& =& \frac{AE}{EC}+\frac{EC}{EC}\\ & & \\ \frac{AF+FC}{FC}& =& \frac{AE+EC}{EC}\end{array}$

Both numerators on both sides of the equation are representations of the segment $AC$. So, we can replace them with $AC$

$\frac{AC}{FC}=\frac{AC}{EC}$

Let's simplify further by multiplying both sides by $\frac{1}{AC}$.

$\begin{array}{rcl}\frac{AC}{FC}\times \frac{1}{AC}& =& \frac{AC}{EC}\times \frac{1}{AC}\\ & & \\ \frac{1}{FC}& =& \frac{1}{EC}\end{array}$

Since they are equal, their reciprocals will also be equal. Therefore,

$\overline{FC}=\overline{EC}$

You should observe that $FC$ and $EC$ are on the same line. If they are on the same line, the only way they can be equal is if both segments start at the same point. This means that point F must be equal to point $E$. It also means that the segment $DE$ is the same as $DF$.

This concludes that $DF$ is indeed parallel to $BC$.

## Proportionality Theorems - Key takeaways

- The basic proportionality theorem states that if a line is drawn parallel to one side of a triangle to intersect the other two sides at distinct points, then the other two sides are divided in the same ratio. The figure below gives a visual representation of the theorem.
- The basic proportionality theorem is also referred to as the triangle proportionality theorem and proportionality segment theorem.
- The converse of the basic proportionality theorem is the reverse of the basic proportionality theorem. The theorem states that if a line is drawn to intersect two sides of a triangle at different points such that it cuts the two sides in the same ratio, then the line is parallel to the third side.

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##### Frequently Asked Questions about Proportionality Theorems

What is proportionality theorem?

Proportionality theorems are for the purpose of showing relationships in form of ratios. They show how different ratios of a figure or a quantity are equal.

How do you solve proportionality theorem?

You don't solve the theorem, you use the theorem to solve problems.

What is the basic proportionality theorem formula?

The basic proportionality theorem formula when there is a triangle ABC with a parallel segment DE is AD/DB = AE/EC

What is the proportionality segment theorem?

The proportionality segment theorem is the same as the basic proportionality theorem.

How can the proportionality theorem be useful to us?

The proportionality theorem is used to show the relationship between the length of sides of triangles. In real life, it can be used in construction.

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