Basic Proportionality Theorem (BPT)

Draw a line inside a triangle parallel to one side, and something quietly exact happens: that line slices the other two sides in exactly the same ratio. This is the Basic Proportionality Theorem, also called Thales' Theorem, and it is one of the two theorems in the CBSE Class 10 chapter Triangles that must be proved, not just applied. It converts a picture — "this line looks parallel" — into an equation you can solve, and it runs equally well in reverse: match a ratio and you have proved a parallel. This article states the theorem and its converse precisely, proves it using the AA similarity criterion built earlier in the chapter, and works through the ratio arithmetic that shows up again and again in board exam questions on triangles.

The core insight

A line parallel to one side of a triangle divides the other two sides in the same ratio, and — read backwards — equal ratios on two sides force the line joining them to be parallel to the third side.

The core idea

Statement. In $\triangle ABC$ , let $D$ be a point on $AB$ and $E$ a point on $AC$ such that $DE \parallel BC$ . Then $D$ and $E$ divide $AB$ and $AC$ in the same ratio:

\frac{AD}{DB} = \frac{AE}{EC}

where: $AD, DB$ are the two parts $DE$ cuts side $AB$ into, and $AE, EC$ are the corresponding parts of side $AC$ .

Why parallel forces proportional. The parallel line $DE$ makes $\triangle ADE$ a scaled-down copy of $\triangle ABC$ — same shape, smaller size — because the two triangles share angle $A$ and $DE \parallel BC$ forces the angle at $D$ to match the angle at $B$ (corresponding angles on the transversal $AB$ ). Two triangles that share all three angles are similar, and similar triangles have proportional sides. That single fact, unpacked carefully below, is the entire theorem. The figure below fixes the labelling that every BPT problem uses.

Figure

Triangle ABC with line DE parallel to BC cutting AB at D and AC at E

Converse. If $D$ is on $AB$ , $E$ is on $AC$ , and $\dfrac{AD}{DB} = \dfrac{AE}{EC}$ , then $DE \parallel BC$ . This is the direction to use when a problem gives you lengths and asks you to prove parallelism, rather than the other way round. The two statements are logical mirror images: the theorem starts from a geometric fact (parallel) and ends at an algebraic one (proportional), while the converse starts from the algebraic fact and ends at the geometric one. Exam answers lose marks precisely when a student states one but proves the other, so always check which direction the question is actually asking for before writing the first line of a solution.

Equivalent forms. $\dfrac{AD}{DB}=\dfrac{AE}{EC}$ can be rewritten with whole sides as $\dfrac{AD}{AB}=\dfrac{AE}{AC}$ , or as $\dfrac{DB}{AB}=\dfrac{EC}{AC}$ . All three say the same thing; pick whichever form matches the lengths you are actually given, since mixing forms is the most common algebra slip.

Derivation AProving AD/DB = AE/EC from parallel lines

The goal is to start from nothing but the given condition $DE \parallel BC$ in $\triangle ABC$ (with $D$ on $AB$ and $E$ on $AC$ ), use the AA similarity criterion already established for this chapter, and arrive at the exact ratio equality $AD/DB = AE/EC$ through algebra alone, with every step justified so the proof can be reproduced on demand in an exam.

Step 1. Compare $\triangle ADE$ and $\triangle ABC$ . Angle $A$ is shared by both triangles (it is literally the same angle). Since $DE \parallel BC$ and $AB$ cuts across both parallel lines as a transversal, the angle at $D$ inside $\triangle ADE$ equals the angle at $B$ inside $\triangle ABC$ — corresponding angles on parallel lines are equal. Two matching angles are enough: by the AA similarity criterion, $\triangle ADE \sim \triangle ABC$ .

Step 2. Convert similarity into a side ratio. Similar triangles have proportional corresponding sides, and $AD$ corresponds to $AB$ while $AE$ corresponds to $AC$ (matching the vertex order $A \to D \to E$ against $A \to B \to C$ ). So $\dfrac{AD}{AB} = \dfrac{AE}{AC}$ . This is a "part-to-whole" ratio, not yet the "part-to-part" form the theorem asks for.

Step 3. Rewrite the whole sides in terms of the parts. Since $D$ lies on $AB$ , $AB = AD + DB$ ; since $E$ lies on $AC$ , $AC = AE + EC$ . Substituting, $\dfrac{AD}{AD+DB} = \dfrac{AE}{AE+EC}$ .

Step 4. Flip both sides (take reciprocals — valid since no side length is zero) to get $\dfrac{AD+DB}{AD} = \dfrac{AE+EC}{AE}$ , which splits as $1 + \dfrac{DB}{AD} = 1 + \dfrac{EC}{AE}$ . Subtracting $1$ from both sides gives $\dfrac{DB}{AD} = \dfrac{EC}{AE}$ , and flipping once more gives the required result, $\dfrac{AD}{DB} = \dfrac{AE}{EC}$ . Verified: with $AD=3, DB=4.5, AE=2.4, EC=3.6$ from the worked example below, $AD/DB = 3/4.5 = 0.667$ and $AE/EC = 2.4/3.6 = 0.667$ , matching to three decimal places as the proof predicts.

Worked example

In $\triangle ABC$ , $D$ and $E$ lie on sides $AB$ and $AC$ such that $DE \parallel BC$ . Given $AD = 3\ \text{cm}$ , $DB = 4.5\ \text{cm}$ , and $AE = 2.4\ \text{cm}$ , find $EC$ and the full length of $AC$ .

Step 1. Since $DE \parallel BC$ , the Basic Proportionality Theorem applies directly: $\dfrac{AD}{DB} = \dfrac{AE}{EC}$ .

Step 2. Substitute the known values: $\dfrac{3}{4.5} = \dfrac{2.4}{EC}$ .

Step 3. Cross-multiply and solve: $3 \times EC = 4.5 \times 2.4 = 10.8$ , so $EC = 10.8/3 = 3.6\ \text{cm}$ .

Step 4. Find $AC$ by adding the parts: $AC = AE + EC = 2.4 + 3.6 = 6\ \text{cm}$ . Answer: $EC = 3.6\ \text{cm}$ and $AC = 6\ \text{cm}$ .

Competitive stretch

In $\triangle PQR$ , $S$ is on $PQ$ with $PS = 4\ \text{cm}$ , $SQ = 6\ \text{cm}$ , and $T$ is on $PR$ with $PT = 6\ \text{cm}$ , $TR = 9\ \text{cm}$ . Is $ST \parallel QR$ ?

Step 1. This asks for a proof of parallelism, so use the converse of BPT: check whether $\dfrac{PS}{SQ} = \dfrac{PT}{TR}$ .

Step 2. Compute both ratios: $\dfrac{PS}{SQ} = \dfrac{4}{6} = \dfrac{2}{3}$ and $\dfrac{PT}{TR} = \dfrac{6}{9} = \dfrac{2}{3}$ .

Step 3. The ratios are equal, so by the converse of the Basic Proportionality Theorem, $ST \parallel QR$ . Answer: yes, $ST$ is parallel to $QR$ — no angle measurement was needed at all, only the ratio check.

Common traps

Mixing "part-to-part" and "part-to-whole" ratios in the same equation — writing $\dfrac{AD}{DB} = \dfrac{AE}{AC}$ instead of keeping both sides in the same form.
Applying the theorem when the line is not actually given (or proved) to be parallel to the third side — the ratio conclusion depends entirely on that condition.
Using the theorem's forward direction ("parallel implies proportional") when the problem actually needs the converse ("proportional implies parallel"), or vice versa.
Forgetting that $D$ and $E$ must lie on the two sides themselves, distinct from the vertices — the theorem does not apply to a line through a vertex or to points on an extended side.
Rushing the converse check by comparing only one pair of ratios visually instead of computing both fractions in lowest terms — two ratios can look close but differ once actually divided out, and the converse only holds when they are exactly equal.

Ask the tutor: If a BPT question mixes whole-side and part-side lengths and you're unsure which ratio form to set up, ask the AI tutor to help you relabel the triangle and pick the cleanest equation before solving.

How is BPT different from just saying the triangles are similar

Similarity is the tool inside the proof, not the final statement. Once you know $\triangle ADE \sim \triangle ABC$ , you get the part-to-whole ratio $AD/AB = AE/AC$ for free, but BPT specifically converts that into the part-to-part ratio $AD/DB = AE/EC$ , which is the form most exam questions actually hand you (segment lengths on either side of the parallel line, not the whole side). Knowing both forms, and how to move between them with the $AB = AD + DB$ substitution, is what the theorem adds beyond bare similarity.

Do I always have to prove BPT from scratch in an exam, or can I just quote it

It depends on the question. If the question says "state" or "using BPT, find," you may quote the ratio directly and use it to solve for the unknown length. If the question says "prove" or asks you to "prove that a line parallel to one side of a triangle divides the other two sides in the same ratio," you must reproduce the AA-similarity argument step by step, since BPT is one of the two proof-required theorems in this CBSE chapter — quoting it without proof loses the marks the question is asking for.

Can BPT be applied to any line inside a triangle, not just one through two sides

No — the theorem strictly needs a line that is parallel to one side and meets the other two sides at two distinct points, both different from the shared vertex. A line through a vertex, a line meeting only one side, or a line that isn't actually parallel to the third side falls outside the theorem's scope, and no proportionality can be assumed for it. Always check the parallel condition is explicitly given or already proved before writing down the ratio.

Basic Proportionality Theorem (BPT)

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads