What is the Basic Proportionality Theorem (Thales Theorem) with statement, proof, and converse?

Direct Answer

The Basic Proportionality Theorem (BPT), also known as Thales' Theorem, states that:

> *"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."*

In mathematical terms, for $\triangle ABC$, if a line parallel to side $BC$ intersects $AB$ at point $D$ and $AC$ at point $E$ (such that $DE \parallel BC$), then:

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

---

Step-by-Step Formal Geometric Proof

#### 1. Given

In $\triangle ABC$, a line $DE$ is drawn parallel to $BC$, intersecting $AB$ at $D$ and $AC$ at $E$.

#### 2. To Prove

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

#### 3. Construction

1. Join vertices $B$ to $E$ and $C$ to $D$ (drawing segments $BE$ and $CD$).

2. Draw perpendicular $DM \perp AC$ from $D$ to side $AC$.

3. Draw perpendicular $EN \perp AB$ from $E$ to side $AB$.

#### 4. Proof Steps

Recall that the area of a triangle is given by:

$$\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$$

Step 1: Compute Area Ratio with base on $AB$

• $\text{Area}(\triangle ADE) = \frac{1}{2} \times AD \times EN$
• $\text{Area}(\triangle BDE) = \frac{1}{2} \times DB \times EN$ *(since $EN$ is the common altitude)*

Dividing these two areas:

$$\frac{\text{Area}(\triangle ADE)}{\text{Area}(\triangle BDE)} = \frac{\frac{1}{2} \times AD \times EN}{\frac{1}{2} \times DB \times EN} = \frac{AD}{DB} \quad \text{--- (Equation 1)}$$

Step 2: Compute Area Ratio with base on $AC$

• $\text{Area}(\triangle ADE) = \frac{1}{2} \times AE \times DM$
• $\text{Area}(\triangle CDE) = \frac{1}{2} \times EC \times DM$

Dividing these two areas:

$$\frac{\text{Area}(\triangle ADE)}{\text{Area}(\triangle CDE)} = \frac{\frac{1}{2} \times AE \times DM}{\frac{1}{2} \times EC \times DM} = \frac{AE}{EC} \quad \text{--- (Equation 2)}$$

Step 3: Equate Denominators

Notice that $\triangle BDE$ and $\triangle CDE$ lie on the same base $DE$ and between the same parallel lines $DE$ and $BC$.

By geometric theorem, triangles on the same base and between the same parallels have equal areas:

$$\text{Area}(\triangle BDE) = \text{Area}(\triangle CDE) \quad \text{--- (Equation 3)}$$

Step 4: Conclusion

From Equations (1), (2), and (3), the left-hand sides are identical. Therefore, their right-hand sides must be equal:

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

*(Hence Proved).*

---

Converse of Basic Proportionality Theorem

> *"If a line divides any two sides of a triangle in the same ratio, then the line must be parallel to the third side."*

If in $\triangle ABC$, $\frac{AD}{DB} = \frac{AE}{EC}$, then $DE \parallel BC$.

---

References & Curriculum

• NCERT Class 10 Mathematics, Chapter 6: *Triangles*, Theorem 6.1.
• CBSE & State Board Secondary Examination Class 10 standard question.