MP Board · Class 10 · Mathematics · TrianglesState and prove the Basic Proportionality Theorem (Thales Theorem).

Statement of Basic Proportionality Theorem (Thales Theorem):\nIf 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.

Proof:

• Given: In $\triangle ABC$, a line $DE$ parallel to side $BC$ intersects $AB$ at $D$ and $AC$ at $E$.
• To Prove: $\frac{AD}{DB} = \frac{AE}{EC}$
• Construction: Join $B$ to $E$ and $C$ to $D$. Draw $DM \perp AC$ and $EN \perp AB$.

Step-by-Step Derivation:

1. Area of Triangles Involving $DE$: The area of $\triangle ADE$ is given by: $$\text{Area}(\triangle ADE) = \frac{1}{2} \times \text{Base} \times \text{Height} = \frac{1}{2} \times AD \times EN$$ Similarly, for $\triangle BDE$ with base $DB$ and height $EN$: $$\text{Area}(\triangle BDE) = \frac{1}{2} \times DB \times EN$$

2. Taking the Ratio: Dividing $\text{Area}(\triangle ADE)$ by $\text{Area}(\triangle BDE)$: $$\frac{\text{Area}(\triangle ADE)}{\text{Area}(\triangle BDE)} = \frac{\frac{1}{2} \times AD \times EN}{\frac{1}{2} \times DB \times EN} = \frac{AD}{DB} \quad \text{--- (Equation 1)}$$

3. Applying Similar Steps for the Other Side: Consider $\triangle ADE$ with base $AE$ and height $DM$: $$\text{Area}(\triangle ADE) = \frac{1}{2} \times AE \times DM$$ And for $\triangle DEC$ with base $EC$ and height $DM$: $$\text{Area}(\triangle DEC) = \frac{1}{2} \times EC \times DM$$

4. Taking the Ratio: Dividing $\text{Area}(\triangle ADE)$ by $\text{Area}(\triangle DEC)$: $$\frac{\text{Area}(\triangle ADE)}{\text{Area}(\triangle DEC)} = \frac{\frac{1}{2} \times AE \times DM}{\frac{1}{2} \times EC \times DM} = \frac{AE}{EC} \quad \text{--- (Equation 2)}$$

5. Relating the Triangles: We know that $\triangle BDE$ and $\triangle DEC$ lie on the same base $DE$ and between the same parallel lines $DE$ and $BC$. Therefore, their areas are equal: $$\text{Area}(\triangle BDE) = \text{Area}(\triangle DEC) \quad \text{--- (Equation 3)}$$

6. Conclusion: From Equation 1, Equation 2, and Equation 3, the left-hand sides are equal. Hence, the right-hand sides must also be equal: $$\frac{AD}{DB} = \frac{AE}{EC}$$ Hence proved.