BPT (Basic Proportionality Theorem) — proof, converse, and applications

Question

What is the Basic Proportionality Theorem (BPT / Thales’ theorem), how do we prove it, and how do we apply it and its converse?

Solution — Step by Step

BPT (Thales’ Theorem): If a line is drawn parallel to one side of a triangle, it divides the other two sides in the same ratio.

In triangle $ABC$ , if $DE \parallel BC$ where $D$ is on $AB$ and $E$ is on $AC$ , then:

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

Draw $BE$ and $CD$ . Drop perpendiculars from $E$ to $AB$ (height $h_1$ ) and from $D$ to $AC$ (height $h_2$ ).

\frac{\text{ar}(\triangle ADE)}{\text{ar}(\triangle BDE)} = \frac{\frac{1}{2} \times AD \times h_1}{\frac{1}{2} \times DB \times h_1} = \frac{AD}{DB} \quad \ldots (1)

\frac{\text{ar}(\triangle ADE)}{\text{ar}(\triangle CDE)} = \frac{\frac{1}{2} \times AE \times h_2}{\frac{1}{2} \times EC \times h_2} = \frac{AE}{EC} \quad \ldots (2)

Now, $\triangle BDE$ and $\triangle CDE$ share the same base $DE$ and lie between the same parallels $DE$ and $BC$ .

\therefore \text{ar}(\triangle BDE) = \text{ar}(\triangle CDE) \quad \ldots (3)

From (1), (2), and (3):

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

If a line divides two sides of a triangle in the same ratio, then the line is parallel to the third side.

If $\frac{AD}{DB} = \frac{AE}{EC}$ , then $DE \parallel BC$ .

This converse is equally important for solving problems — it lets us conclude that lines are parallel.

Problem: In $\triangle ABC$ , $D$ is on $AB$ and $E$ is on $AC$ such that $DE \parallel BC$ . If $AD = 4$ cm, $DB = 6$ cm, and $AE = 3$ cm, find $EC$ .

By BPT: $\frac{AD}{DB} = \frac{AE}{EC}$

\frac{4}{6} = \frac{3}{EC}

EC = \frac{6 \times 3}{4} = 4.5 \text{ cm}

flowchart TD
    A["BPT Problem"] --> B{"What is given?"}
    B -->|"DE parallel to BC + side lengths"| C["Use BPT: AD/DB = AE/EC"]
    B -->|"Ratio of sides equal"| D["Use converse: conclude DE parallel to BC"]
    C --> E["Set up proportion, cross multiply, solve"]
    D --> F["State the converse theorem"]
    A --> G["Also useful: AD/AB = AE/AC from BPT"]

Why This Works

The proof is elegant because it connects proportionality to area. Two triangles with the same height have areas proportional to their bases. And two triangles between the same parallels with the same base have equal areas. Combining these two area properties gives us the proportionality result.

BPT is the geometric foundation for the concept of similarity. In fact, the AA similarity criterion can be derived from BPT.

Alternative Method

An equivalent form of BPT that is sometimes more convenient:

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

This version directly gives the ratio of the parallel line segment $DE$ to the side $BC$ , which is useful when the problem asks for the length of $DE$ .

Common Mistake

Students write $\frac{AD}{DB} = \frac{AE}{EC}$ as $\frac{AD}{DB} = \frac{AE}{AC}$ (using $AC$ instead of $EC$ ). The ratio is between the two parts of each side, not between one part and the whole side. If $AD/DB = AE/EC$ , then $AD/AB = AE/AC$ is also true (componendo), but the original form has parts over parts, not part over whole. Mixing these up changes the calculation completely. This costs marks in CBSE 10th proofs and numericals alike.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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