Mathematics — Standard (041) — CBSE Class 10 board question

Statement (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 is on AB and E is on AC.

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

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

Proof:

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

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

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):

$$\boxed{\frac{AD}{DB} = \frac{AE}{EC}}$$ [Proved]

Source: Theorem 6.1, Chapter 6 — Triangles

---

• Statement must be written word-for-word (1 mark).
• Given / To Prove / Construction each fetch marks — don't skip them.
• The proof hinges on equal areas of △BDE and △DEC (same base DE, same parallels BC ∥ DE); this is the key step examiners check.
• Write equations numbered (1), (2), (3) and explicitly conclude from all three — examiners award the final mark only when the logical chain is complete.
• A neat labelled diagram (△ABC with D on AB, E on AC, DE ∥ BC) adds presentation marks.