The basic proportionality theorem

A line parallel to a side

Take a triangle $ABC$. Draw a line $DE$ parallel to the side $BC$, with $D$ on $AB$ and $E$ on $AC$. The line cuts $AB$ and $AC$ into pieces. More than two thousand five hundred years ago, the Greek mathematician Thales noticed that the pieces are in proportion.

The basic proportionality theorem

Theorem. Draw a line parallel to one side of a triangle, so that it meets the other two sides at two different points. Then it divides those two sides in the same ratio.

In triangle $ABC$, if $DE \parallel BC$, then

Why it is true

We prove it with areas. We use one fact: triangles with the same height have areas in the ratio of their bases.

1. Join $B$ to $E$ and $C$ to $D$.
2. Triangles $ADE$ and $BDE$ have the same height, the distance from $E$ to the line $AB$. So $\dfrac{\text{ar}(ADE)}{\text{ar}(BDE)} = \dfrac{AD}{DB}$.
3. Triangles $ADE$ and $CDE$ have the same height, the distance from $D$ to the line $AC$. So $\dfrac{\text{ar}(ADE)}{\text{ar}(CDE)} = \dfrac{AE}{EC}$.
4. Triangles $BDE$ and $CDE$ stand on the same base $DE$ and lie between the same parallel lines $DE$ and $BC$. So $\text{ar}(BDE) = \text{ar}(CDE)$.
5. By step 4, the bottom lines in steps 2 and 3 are equal. So the two ratios are equal: $\dfrac{AD}{DB} = \dfrac{AE}{EC}$.

Step 4 is the only step that uses the parallel lines. That is why the theorem needs $DE \parallel BC$.

Finding a missing length

In triangle $ABC$, $DE \parallel BC$, with $AD = 3$ cm, $DB = 5$ cm and $AE = 4.5$ cm. Find $EC$.

1. Write the theorem. $\dfrac{AD}{DB} = \dfrac{AE}{EC}$.
2. Put in the lengths. $\dfrac{3}{5} = \dfrac{4.5}{EC}$.
3. Solve. $EC = \dfrac{4.5 \times 5}{3} = 7.5$ cm.

Check. $\dfrac{4.5}{7.5} = 0.6$ and $\dfrac{3}{5} = 0.6$. Both ratios agree.

When the lengths are algebra

Suppose $DE \parallel BC$, with $AD = x$ cm, $DB = 2$ cm, $AE = (x + 1)$ cm and $EC = 3$ cm. Find $x$.

So $AD = 2$ cm and $AE = 3$ cm.

Other forms of the same fact

Add $1$ to both sides of $\dfrac{AD}{DB} = \dfrac{AE}{EC}$. Then $\dfrac{AD + DB}{DB} = \dfrac{AE + EC}{EC}$, which says $\dfrac{AB}{DB} = \dfrac{AC}{EC}$.

Turn the ratios upside down and add $1$ in the same way, and we get a third form:

These are all the same fact. Use the form that puts the unknown in the easiest place. For example, if $AB = 10$ cm, $AD = 4$ cm and $AC = 15$ cm, then $AE = \dfrac{AD}{AB} \times AC = \dfrac{4}{10} \times 15 = 6$ cm.

Practise with new numbers

Try it in Python

The program places $D$ and $E$ by coordinates, for three positions, and checks both ratios and the parallel lines.

Common mistakes

• Pairing the wrong lengths. $\dfrac{AD}{DB} = \dfrac{AE}{EC}$ compares a piece with a piece. $\dfrac{AD}{DB} = \dfrac{AE}{AC}$ compares a piece with the whole, which is wrong.
• Turning only one ratio upside down. If you flip one side, flip the other.
• Using the theorem when the lines are not parallel. The theorem needs $DE \parallel BC$. Without it the ratios are not equal.
• Dropping the units. The lengths are in centimetres, and the ratios have no unit.

Remember these

• If $DE \parallel BC$, then $\dfrac{AD}{DB} = \dfrac{AE}{EC}$.
• The same fact can be written $\dfrac{AB}{DB} = \dfrac{AC}{EC}$ or $\dfrac{AD}{AB} = \dfrac{AE}{AC}$.
• The proof compares areas of triangles with equal heights, and uses the parallel lines for equal areas.
• A piece goes with a piece, and a whole goes with a whole.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. In triangle $ABC$, $DE \parallel BC$, with $AD = 6$ cm, $DB = 9$ cm and $AE = 4$ cm. Find $EC$.

2. With the same triangle, find $AC$.

3. In triangle $PQR$, $ST \parallel QR$, with $S$ on $PQ$ and $T$ on $PR$. $PQ = 12$ cm, $PS = 3$ cm and $PR = 20$ cm. Find $PT$.

4. Neel writes $\dfrac{AD}{DB} = \dfrac{AE}{AC}$. What is wrong with it?

5. Which step of the proof needs $DE$ to be parallel to $BC$?