Mathematics — Standard (041) — 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 at distinct points, then the other two sides are divided in the same ratio.

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

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

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

Proof:

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

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

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

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

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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• Marks split: ~1 mark for Given/To Prove/Construction, ~4 marks for the proof steps.
• The key idea is expressing both ratios as areas of triangles sharing the same height, then using the fact that △BDE and △DEC are equal in area (same base DE, same parallels BC ∥ DE).
• Always write the construction clearly — joining BE, CD and drawing perpendiculars is essential to set up the area argument.
• Do not skip equations (1), (2), (3) — examiners look for all three steps explicitly stated.