What Does the Basic Proportionality Theorem State?
The Basic Proportionality Theorem states: if a line is drawn parallel to one side of a triangle to intersect the other two sides at distinct points, it divides those two sides in the same ratio. In triangle $ABC$, if a line $DE$ is drawn parallel to side $BC$, meeting $AB$ at $D$ and $AC$ at $E$, then:
$$\frac{AD}{DB} = \frac{AE}{EC}$$
Each symbol names a segment of a side: $AD$ and $DB$ are the two pieces of side $AB$, while $AE$ and $EC$ are the two pieces of side $AC$. The theorem promises those two ratios are equal, the parallel line cuts both sides in step.
The Man Who Measured a Pyramid With Its Shadow
Around 600 BCE, Thales measured the height of the Great Pyramid without climbing a single step. He waited until his own shadow equalled his height, then measured the pyramid's shadow, because at that instant, the pyramid's shadow equalled its height too. Behind that trick sits a single idea about proportion inside triangles, and it is the idea this whole theorem makes exact.
How Do You Prove the Basic Proportionality Theorem?
The cleanest proof compares areas. It is worth following once, because it explains why the ratio holds rather than asking you to take it on trust.
Given: triangle $ABC$ with $DE \parallel BC$, where $D$ lies on $AB$ and $E$ lies on $AC$.
To prove: $\dfrac{AD}{DB} = \dfrac{AE}{EC}$.
Construction: join $BE$ and $CD$. Drop perpendiculars $EN$ from $E$ onto $AB$, and $DM$ from $D$ onto $AC$.
Proof steps:
Triangles $ADE$ and $BDE$ share the same height $EN$ (the perpendicular from $E$), with bases $AD$ and $DB$. So $$\frac{\text{ar}(ADE)}{\text{ar}(BDE)} = \frac{AD}{DB}$$
Triangles $ADE$ and $CDE$ share the same height $DM$, with bases $AE$ and $EC$. So $$\frac{\text{ar}(ADE)}{\text{ar}(CDE)} = \frac{AE}{EC}$$
Triangles $BDE$ and $CDE$ sit on the same base $DE$ and between the same parallels $DE$ and $BC$, so they have equal area: $$\text{ar}(BDE) = \text{ar}(CDE)$$
The left-hand sides of steps 1 and 2 therefore have equal denominators, so $$\frac{AD}{DB} = \frac{AE}{EC} \qquad \blacksquare$$
What Is the Converse of the Basic Proportionality Theorem?
The converse runs the logic backwards and is just as useful: if a line divides two sides of a triangle in the same ratio, then that line is parallel to the third side. So if $\dfrac{AD}{DB} = \dfrac{AE}{EC}$, you may conclude $DE \parallel BC$. The theorem tells you the ratio from the parallel; the converse lets you prove the parallel from the ratio.
Examples of the Basic Proportionality Theorem
The examples move from a one-step ratio to an algebraic solve and a real triangular plot.
Example 1
In triangle $ABC$, $DE \parallel BC$ with $D$ on $AB$ and $E$ on $AC$. If $AD = 3$ cm, $DB = 6$ cm, and $AE = 4$ cm, find $EC$.
Apply the theorem directly.
$$\frac{AD}{DB} = \frac{AE}{EC}$$
$$\frac{3}{6} = \frac{4}{EC}$$
$$EC = \frac{4 \times 6}{3} = 8 \text{ cm}$$
Final answer: $EC = 8$ cm.
Example 2
In triangle $ABC$, $DE \parallel BC$. If $AD = 4$, $DB = 6$, and $AE = 6$, find $EC$.
A frequent first attempt matches $AD : DB$ against $AE : AC$, pairing a piece of one side with the whole of the other. That gives $\frac{4}{6} = \frac{6}{AC}$, so $AC = 9$ and $EC = AC - AE = 9 - 6 = 3$. Look at that result against the diagram: $EC = 3$ would make $E$ sit far up side $AC$, yet $AE = 6$ is already the larger piece, so $EC$ cannot be the smaller one if the cut is proportional to $AD < DB$. The setup mismatched a segment with a whole side.
The theorem pairs segment with corresponding segment: $AD : DB = AE : EC$.
$$\frac{4}{6} = \frac{6}{EC}$$
$$EC = \frac{6 \times 6}{4} = 9$$
Final answer: $EC = 9$, larger than $AE$, matching $DB > AD$, exactly as the proportion requires.
Example 3
In triangle $PQR$, $ST \parallel QR$ with $S$ on $PQ$ and $T$ on $PR$. If $PS = 5$ cm, $SQ = 10$ cm, and $PT = 7.5$ cm, find $TR$.
$$\frac{PS}{SQ} = \frac{PT}{TR}$$
$$\frac{5}{10} = \frac{7.5}{TR}$$
$$TR = \frac{7.5 \times 10}{5} = 15 \text{ cm}$$
Final answer: $TR = 15$ cm.
Example 4
In triangle $ABC$, $D$ lies on $AB$ and $E$ on $AC$ with $AD = 2$, $DB = 3$, $AE = 4$, $EC = 6$. Is $DE$ parallel to $BC$?
This calls for the converse, compute both ratios and compare.
$$\frac{AD}{DB} = \frac{2}{3}, \qquad \frac{AE}{EC} = \frac{4}{6} = \frac{2}{3}$$
The ratios are equal, so by the converse of the theorem, $DE \parallel BC$.
Final answer: yes, $DE$ is parallel to $BC$.
Example 5
In triangle $ABC$, $DE \parallel BC$ with $AD = x$, $DB = 4$, $AE = x + 1$, and $EC = 6$. Find $x$.
Set up the proportion and solve.
$$\frac{x}{4} = \frac{x + 1}{6}$$
$$6x = 4(x + 1)$$
$$6x = 4x + 4$$
$$2x = 4 \quad\Rightarrow\quad x = 2$$
Final answer: $x = 2$.
Example 6
A triangular garden $ABC$ has a straight path $DE$ laid parallel to the base $BC$, with $D$ the midpoint of side $AB$. If $AB = 12$ m and $AC = 10$ m, where does the path meet side $AC$?
Since $D$ is the midpoint, $AD = DB = 6$ m, so $\dfrac{AD}{DB} = 1$. By the theorem, $\dfrac{AE}{EC} = 1$ as well.
$$AE = EC = \frac{AC}{2} = \frac{10}{2} = 5 \text{ m}$$
Final answer: the path meets $AC$ at its midpoint, $5$ m from $A$. This is exactly the midpoint theorem, a special case of BPT when the parallel line starts at a midpoint.
Why Does the Basic Proportionality Theorem Matter?
The theorem is the foundation of similarity, the whole idea that shapes can share a form at different sizes. Once you know a parallel line splits two sides proportionally, you can prove that the small triangle $ADE$ and the full triangle $ABC$ are similar triangles, which unlocks height-and-distance problems, scale drawings, and map work.
Measuring the unreachable: Thales' pyramid trick, and every shadow-and-stick height measurement since, rests on proportional triangles. The MacTutor biography of Thales of Miletus records the method.
The midpoint theorem: as Example 6 shows, a parallel through a midpoint hits the opposite midpoint, a direct corollary.
Proportion in general: BPT is a geometric statement of ordinary proportion, the same $a : b = c : d$ reasoning used across arithmetic.
Where Do You Actually Use the Converse?
Any time you need to prove two lines are parallel from measurements alone, in a construction, a proof, or a survey, you check the ratio and invoke the converse. It turns a numerical fact into a geometric conclusion without a protractor.
What Are the Most Common Mistakes With BPT?
Mistake 1: Matching a segment against the whole side
Where it slips in: setting up the proportion, especially when the diagram is drawn quickly.
Don't do this: write $\dfrac{AD}{DB} = \dfrac{AE}{AC}$, pairing a piece of one side with the entire other side.
The correct way: pair each segment with its corresponding segment: $\dfrac{AD}{DB} = \dfrac{AE}{EC}$. The reliable error is lining up a piece against a whole; keep both ratios piece-against-piece.
Mistake 2: Applying the theorem without a genuine parallel line
Where it slips in: problems where $DE$ is drawn across the triangle but is not stated or proven to be parallel to $BC$.
Don't do this: assume the proportion holds for any cutting line.
The correct way: the equal-ratio conclusion needs $DE \parallel BC$. Without the parallel condition, use the converse only to test for parallelism, not to assume it.
Mistake 3: Flipping one ratio but not the other
Where it slips in: cross-multiplying in a hurry.
Don't do this: write $\dfrac{AD}{DB} = \dfrac{EC}{AE}$, inverting one side.
The correct way: keep both fractions in the same order, top-of-$AB$ over bottom-of-$AB$ equals top-of-$AC$ over bottom-of-$AC$.
Conclusion
The Basic Proportionality Theorem says a line parallel to one side of a triangle divides the other two sides in equal ratio: $\dfrac{AD}{DB} = \dfrac{AE}{EC}$.
The proof compares triangle areas that share a base between the same parallels.
The converse lets you prove a line is parallel from equal ratios alone.
BPT is the gateway to similar triangles and the midpoint theorem, and to every shadow-and-proportion measurement since Thales.
To build proportion and similarity with a teacher, explore Bhanzu's geometry tutor or high school math tutor, or join a live math class online.
A Practical Next Step
Practice these problems to solidify your understanding: redo Example 2 using the equivalent form $\dfrac{AD}{AB} = \dfrac{AE}{AC}$ and confirm you reach the same $EC$. If the ratios feel slippery, return to Mistake 1. See how BPT extends into similar figures next. Want a live trainer to check your proofs? Book a free demo class.
Read More
Types of triangles - how triangles are classified by their sides and angles.
Areas of similar triangles - why areas scale with the square of the side ratio.
Scale factor - the number that links corresponding sides of proportional figures.
Angle bisector theorem - another line that divides a triangle's side in a fixed ratio.
Congruence in triangles - how equal-size triangles differ from merely similar ones.
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