CBSE · Class 10 · Mathematics · TrianglesState and prove the Basic Proportionality Theorem (Thales's Theorem). Also, in triangle ABC, DE is parallel to BC intersecting AB at D and AC at E. If AD = 1.5 cm, DB = 3 cm, and AE = 1 cm, find the length of EC.

Statement of Thales's 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 $\triangle ABC$, $DE \parallel BC$, intersecting $AB$ 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 $EN \perp AB$.

Proof Steps:

1. The area of $\triangle ADE$ is given by: $$\text{Area}(\triangle ADE) = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times AD \times EN$$
2. The area of $\triangle BDE$ is: $$\text{Area}(\triangle BDE) = \frac{1}{2} \times DB \times EN$$
3. Dividing the two areas: $$\frac{\text{Area}(\triangle ADE)}{\text{Area}(\triangle BDE)} = \frac{\frac{1}{2} \times AD \times EN}{\frac{1}{2} \times DB \times EN} = \frac{AD}{DB} \quad \text{--- (Equation 1)}$$
4. Similarly, considering $AE$ and $EC$ as bases with height $DM$: $$\frac{\text{Area}(\triangle ADE)}{\text{Area}(\triangle DEC)} = \frac{\frac{1}{2} \times AE \times DM}{\frac{1}{2} \times EC \times DM} = \frac{AE}{EC} \quad \text{--- (Equation 2)}$$
5. Since $\triangle BDE$ and $\triangle DEC$ lie on the same base $DE$ and between the same parallel lines $DE$ and $BC$, their areas are equal: $$\text{Area}(\triangle BDE) = \text{Area}(\triangle DEC) \quad \text{--- (Equation 3)}$$
6. From Equations (1), (2), and (3), we conclude that: $$\frac{AD}{DB} = \frac{AE}{EC}$$\nHence proved.

Numerical Solution:

Given data:

• $AD = 1.5\text{ cm}$
• $DB = 3\text{ cm}$
• $AE = 1\text{ cm}$
• $EC = ?$

Formula:\nAccording to Thales's Theorem: $$\frac{AD}{DB} = \frac{AE}{EC}$$

Substitution: $$\frac{1.5}{3} = \frac{1}{EC}$$

Calculation: $$1.5 \times EC = 3 \times 1$$ $$1.5 \times EC = 3$$ $$EC = \frac{3}{1.5}$$ $$EC = 2\text{ cm}$$

Answer:\nThe length of $EC$ is $2\text{ cm}$.