Mathematics — CBSE Class 10 board question

Theorem (Basic Proportionality Theorem / Thales Theorem): If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio.

Given: In △ABC, DE ∥ BC, where D lies on AB and E lies on AC.

To Prove: $\dfrac{AD}{DB} = \dfrac{AE}{EC}$

Construction: Join BE and CD. Draw DM ⊥ AC and EN ⊥ AB.

Proof:

$$\frac{\text{ar(ADE)}}{\text{ar(BDE)}} = \frac{\frac{1}{2} \times AD \times EN}{\frac{1}{2} \times DB \times EN} = \frac{AD}{DB} \quad \cdots(1)$$

$$\frac{\text{ar(ADE)}}{\text{ar(DEC)}} = \frac{\frac{1}{2} \times AE \times DM}{\frac{1}{2} \times EC \times DM} = \frac{AE}{EC} \quad \cdots(2)$$

Since △BDE and △DEC lie on the same base DE and between the same parallels BC and DE:

$$\text{ar(BDE)} = \text{ar(DEC)} \quad \cdots(3)$$

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

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

Hence proved.

Source: Chapter 6, Section 6.3 — Theorem 6.1

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• What examiners look for: Clear statement of Given/To Prove/Construction/Proof — all four parts are expected for a 5-mark proof. Missing any part costs marks.
• Key construction: Drawing DM ⊥ AC and EN ⊥ AB is essential to express the areas of the triangles using different bases.
• Critical step: The fact that △BDE and △DEC have equal areas (same base DE, same parallels BC ∥ DE) is the pivot of the proof — don't skip the reason.
• Common mistake: Students forget to state that ar(BDE) = ar(DEC) with a reason, or fail to link equations (1), (2), and (3) explicitly.
• This theorem is also called the Basic Proportionality Theorem (BPT) or Thales Theorem — either name is acceptable.