Proportionality Theorems

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 figure below gives a visual representation of the theorem.

A Triangle - StudySmarter Originals

In the $∆ABC$ 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 $EC$:

The ratio above is considered the basic proportionality formula.

The Triangle Proportionality Theorem and Fundamental Theorem of Proportionality

The Triangle 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.

Let's take a look at another example.

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.

A Triangle - StudySmarter Originals

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

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.

A triangle divided into parts with segments - StudySmarter Originals

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.

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 $BC$.