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

Question

In triangle ABC, a line DE is drawn parallel to BC such that D lies on AB and E lies on AC. If AD = 4 cm, DB = 6 cm, and AE = 3 cm, find EC. State and prove the Basic Proportionality Theorem.

(CBSE Class 10 pattern)

Solution — Step by Step

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 $DE \parallel BC$ , then:

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

flowchart TD
    A["DE ∥ BC in △ABC"] --> B["BPT applies"]
    B --> C["AD/DB = AE/EC"]
    B --> D["Equivalent forms:\nAD/AB = AE/AC\nDB/AB = EC/AC"]
    C --> E["Converse also true:\nIf AD/DB = AE/EC\nthen DE ∥ BC"]

Given: $AD = 4$ , $DB = 6$ , $AE = 3$ , $EC = ?$

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

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

EC = \frac{3 \times 6}{4} = \frac{18}{4} = \mathbf{4.5 \text{ cm}}

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

To prove: $AD/DB = AE/EC$

Construction: Draw $EM \perp AB$ and $DN \perp AC$ . Join BE and CD.

Proof:

$\text{Area of } \triangle ADE = \frac{1}{2} \times AD \times EM$ … (using base AD, height EM)

$\text{Area of } \triangle BDE = \frac{1}{2} \times DB \times EM$ … (using base DB, same height EM)

So: $\frac{\text{ar}(\triangle ADE)}{\text{ar}(\triangle BDE)} = \frac{AD}{DB}$ … (i)

Similarly: $\frac{\text{ar}(\triangle ADE)}{\text{ar}(\triangle CDE)} = \frac{AE}{EC}$ … (ii)

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

Therefore: $\text{ar}(\triangle BDE) = \text{ar}(\triangle CDE)$ … (iii)

From (i), (ii), and (iii): $\frac{AD}{DB} = \frac{AE}{EC}$ . Hence proved.

Why This Works

The proof cleverly uses the fact that triangles with the same base between parallel lines have equal areas. This connects the ratio of sides to the ratio of areas, and the equal-area property (from parallelism) completes the chain. No coordinate geometry or complex algebra is needed — just pure area reasoning.

The converse is equally useful: if a line divides two sides of a triangle in equal ratios, then the line must be parallel to the third side. This is often used to prove that lines are parallel.

Alternative Method — Using Similar Triangles

Once BPT is established, we know $\triangle ADE \sim \triangle ABC$ (by AA similarity, since $\angle A$ is common and $\angle ADE = \angle ABC$ as corresponding angles with $DE \parallel BC$ ).

By similarity: $\frac{AD}{AB} = \frac{AE}{AC} = \frac{DE}{BC}$

This gives us the ratio of all corresponding sides, including $DE/BC$ .

In CBSE Class 10 board exams, BPT proof is a guaranteed 5-mark question. The construction step (drawing perpendiculars) is what students forget. Without the perpendiculars, you cannot express the areas in terms of the sides. Practice the proof until the construction comes automatically.

Common Mistake

Students often write $\frac{AD}{DB} = \frac{AE}{AC}$ instead of $\frac{AD}{DB} = \frac{AE}{EC}$ . Notice the denominator: it is $EC$ (the other segment on the same side), not $AC$ (the whole side). While $\frac{AD}{AB} = \frac{AE}{AC}$ is also valid, mixing the forms ( $AD/DB$ on one side and $AE/AC$ on the other) is incorrect. Be consistent: either use segment ratios ( $AD/DB = AE/EC$ ) or whole-side ratios ( $AD/AB = AE/AC$ ).

Question

Solution — Step by Step

Why This Works

Alternative Method — Using Similar Triangles

Common Mistake

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