Basic proportionality theorem

Draw any triangle, then draw a second line inside it parallel to one of the sides. That single parallel line does something remarkable to the two sides it crosses: it always cuts them in exactly the same ratio, no matter how the triangle is shaped or how the line is angled. This is the Basic Proportionality Theorem (BPT), also known as Thales' theorem, and it is one of the most heavily used results in the whole Similarity chapter — every time an ICSE geometry problem hands you a triangle with a parallel line drawn across it, BPT is almost certainly the tool. This article rebuilds BPT from the similar-triangles idea you already know, proves it properly using areas of triangles with equal height, and then drills both the direct theorem (finding an unknown length) and its converse (proving parallelism from given ratios) — the two skills ICSE actually examines.

The core insight

A line parallel to one side of a triangle divides the other two sides in the same ratio, and — by the converse — a line that divides two sides in the same ratio must itself be parallel to the third side.

The core idea

The statement. In $\triangle ABC$ , let $D$ lie on $AB$ and $E$ lie on $AC$ , with $DE \parallel BC$ . Then:

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

where: $AD$ and $AE$ are the segments nearer vertex $A$ ("near" parts), $DB$ and $EC$ are the segments nearer the base ("far" parts), and the two ratios compare corresponding near-to-far pairs on each side. The same relation can also be written part-to-whole:

\frac{AD}{AB} = \frac{AE}{AC}

Why the parallel line matters. As soon as $DE \parallel BC$ , the corresponding angles formed by the transversal sides $AB$ and $AC$ match up: $\angle ADE = \angle ABC$ and $\angle AED = \angle ACB$ , and $\angle A$ is shared by both triangles. That is AA similarity, so $\triangle ADE \sim \triangle ABC$ , and BPT's ratio is really just one consequence of that similarity written out for a single cevian instead of the whole triangle.

The converse. If $D$ is on $AB$ , $E$ is on $AC$ , and $\dfrac{AD}{DB} = \dfrac{AE}{EC}$ , then $DE \parallel BC$ . This direction is used constantly to prove a line is parallel to a side purely from ratio data, with no angle information given at all.

A related picture. The same ratio idea reappears whenever several parallel lines cross two transversals — the intercepts cut on the transversals are proportional, exactly as if one of the transversals were a triangle side split by a line parallel to the third side. Recognising this "same theorem, different picture" version helps with trapezium and ladder-style diagrams.

Figure

Triangle ABC with a line DE parallel to BC cutting AB at D and AC at E, marking the near segments AD and AE and the far segments DB and EC as equal ratios

Derivation AProving the Basic Proportionality Theorem from equal-height triangle areas

The goal is to show, using only the area formula for a triangle and the fact that triangles sharing the same height have areas in the ratio of their bases, that a line $DE$ drawn parallel to side $BC$ of $\triangle ABC$ must divide the other two sides $AB$ and $AC$ in exactly the same ratio, without assuming similarity or angle-chasing as a starting point.

Step 1. Join $BE$ and $CD$ to create four small triangles inside $\triangle ABC$ : $\triangle ADE$ , $\triangle BDE$ , $\triangle CDE$ , and the whole triangle can be read off from these pieces. Compare $\triangle ADE$ and $\triangle BDE$ first — they share the same vertex $E$ , and their bases $AD$ and $DB$ lie on the same straight line $AB$ , so they share a common height (the perpendicular distance from $E$ to line $AB$ ).

Step 2. Write the ratio of their areas using that shared height $h$ . Since area equals half base times height, $\dfrac{\text{ar}(\triangle ADE)}{\text{ar}(\triangle BDE)} = \dfrac{\frac{1}{2}\cdot AD \cdot h}{\frac{1}{2}\cdot DB \cdot h} = \dfrac{AD}{DB}$ . The height cancels because it is identical for both triangles, leaving a pure ratio of bases.

Step 3. Repeat the same argument on the other side. Compare $\triangle ADE$ and $\triangle CDE$ : they share vertex $D$ , and their bases $AE$ and $EC$ lie on line $AC$ , so they share the height from $D$ to line $AC$ . By the same reasoning, $\dfrac{\text{ar}(\triangle ADE)}{\text{ar}(\triangle CDE)} = \dfrac{AE}{EC}$ .

Step 4. Show $\text{ar}(\triangle BDE) = \text{ar}(\triangle CDE)$ , then combine. Both $\triangle BDE$ and $\triangle CDE$ sit on the same base $DE$ and, because $DE \parallel BC$ , both have the same perpendicular distance from $DE$ to line $BC$ — so their areas are equal. Substituting this equality into the two ratios from Steps 2 and 3, both of which have $\text{ar}(\triangle ADE)$ on top and now equal denominators, gives $\dfrac{AD}{DB} = \dfrac{\text{ar}(\triangle ADE)}{\text{ar}(\triangle BDE)} = \dfrac{\text{ar}(\triangle ADE)}{\text{ar}(\triangle CDE)} = \dfrac{AE}{EC}$ , which is exactly the Basic Proportionality Theorem.

Worked example

Question: In $\triangle ABC$ , $D$ and $E$ lie on $AB$ and $AC$ such that $DE \parallel BC$ , with $AD = 3\ \text{cm}$ , $DB = 4.5\ \text{cm}$ , $AE = 4\ \text{cm}$ . Find $EC$ . Then, using $\triangle PQR$ with $S$ on $PQ$ , $T$ on $PR$ , $PS = 4\ \text{cm}$ , $SQ = 6\ \text{cm}$ , $PT = 6\ \text{cm}$ , $TR = 9\ \text{cm}$ , decide whether $ST \parallel QR$ .

Step 1. Apply BPT directly to the first triangle. Since $DE \parallel BC$ , $\dfrac{AD}{DB} = \dfrac{AE}{EC}$ , so $\dfrac{3}{4.5} = \dfrac{4}{EC}$ .

Step 2. Solve for $EC$ . Cross-multiplying, $EC = \dfrac{4 \times 4.5}{3} = \dfrac{18}{3} = 6\ \text{cm}$ .

Step 3. For the second triangle, compute both ratios separately since parallelism is not given. $\dfrac{PS}{SQ} = \dfrac{4}{6} = \dfrac{2}{3}$ and $\dfrac{PT}{TR} = \dfrac{6}{9} = \dfrac{2}{3}$ .

Step 4. Compare and apply the converse. Both ratios equal $\dfrac{2}{3}$ , so by the converse of BPT, $ST \parallel QR$ . Because $PS:PQ = 4:10 = 2:5$ , triangle $PST$ is similar to triangle $PQR$ by AA, so every other pair of corresponding lengths — including $ST$ compared with $QR$ — is also in the ratio $2:5$ . Answer: $EC = 6\ \text{cm}$ , and $ST \parallel QR$ because the two ratios are equal.

Common traps

Mixing "part-to-part" with "part-to-whole" — comparing $\dfrac{AD}{DB}$ on one side with $\dfrac{AC}{AE}$ on the other instead of keeping the same style of ratio on both sides.
Forgetting the converse exists — many exam questions give three or four side lengths and ask you to prove a line is parallel, which needs the converse, not the direct ratio statement.
Applying BPT without checking that a line really is parallel to the third side — if no parallelism is stated or provable, the ratio relation simply does not hold.
Swapping which segment is "near" the vertex versus "far" from it when a diagram is unlabelled or drawn small, leading to an inverted ratio.

Ask the tutor: Want to practise a mixed set of direct-theorem and converse-of-BPT problems, including one where the parallel line meets the extended sides? Ask the AI tutor for extra worked examples.

Is BPT the same as saying the two triangles are similar?

Not quite the same statement, but very closely linked. BPT is about ratios of segments on the two sides cut by the parallel line, while similarity is about the whole shape of $\triangle ADE$ matching $\triangle ABC$ . In fact $DE \parallel BC$ gives you both at once: the equal corresponding angles prove $\triangle ADE \sim \triangle ABC$ by AA, and that similarity is exactly what makes the BPT ratio true. So BPT can be seen as similarity applied to just one cevian inside the triangle.

Do I need to prove BPT from scratch in an ICSE exam, or can I just quote it?

For ICSE Class 10, you are expected to apply BPT and its converse confidently in numeric and reasoning problems — quoting the theorem by name and using the ratio is standard practice and does not need re-proving each time. The area-based proof shown here is worth understanding once so the theorem feels earned rather than memorised, but a full step-by-step proof is not usually demanded unless a question explicitly asks you to prove BPT.

What if the parallel line meets the sides extended beyond the triangle, not between A and the base?

The same underlying ratio relation still holds, with a small adjustment for which segments count as "outside" the triangle rather than inside it. This variant is tested far less often than the standard case with $D$ between $A$ and $B$ and $E$ between $A$ and $C$ , but it is worth recognising the picture if it appears, since the reasoning (equal corresponding angles from the parallel line) is identical.

Basic proportionality theorem

The core idea

Worked example

Common traps

Before this

Where this leads