Class 10 Maths Thales Theorem 6.1

Thales’ Theorem (Basic Proportionality Theorem)

In NCERT Class 10 (Chapter: Triangles), Thales’ Theorem—also called the Basic Proportionality Theorem (BPT)—is a key result used in many geometry problems.

Statement (BPT): If a line is drawn parallel to one side of a triangle and it intersects the other two sides, then it divides those two sides in the same ratio.

In \( \triangle ABC \), let \( D \) be on \( AB \) and \( E \) be on \( AC \). If \( DE \parallel BC \), then \[ \frac{AD}{DB} = \frac{AE}{EC}. \]

📺 Watch: Thales' Theorem explained with proof and examples.

A B C D E
Figure: In \( \triangle ABC \), \( DE\parallel BC \).

Proof (Method 1: Using Areas of Triangles)

  1. Since \( DE \parallel BC \), \( \triangle ADE \) and \( \triangle DBE \) are on the same base \( DE \) and between the same parallels \( DE \parallel BC \). Hence, \[ \text{ar}(\triangle ADE) = \text{ar}(\triangle DBE). \]
  2. Similarly, \( \triangle ADE \) and \( \triangle EDC \) are on the same base \( AE \) and between the same parallels, so \[ \text{ar}(\triangle ADE) = \text{ar}(\triangle EDC). \]
  3. Now, \[ \frac{AD}{DB} = \frac{\text{ar}(\triangle ADE)}{\text{ar}(\triangle DBE)}, \quad \frac{AE}{EC} = \frac{\text{ar}(\triangle ADE)}{\text{ar}(\triangle EDC)}. \]
  4. Since the two ratios are equal, we conclude \[ \frac{AD}{DB} = \frac{AE}{EC}. \]

Hence proved (by areas).

Proof (Method 2: By Similarity of Triangles)

  1. Given \( DE \parallel BC \), we get \( \angle ADE = \angle ABC \) and \( \angle AED = \angle ACB \) (alternate interior angles).
  2. Also, \( \angle A = \angle A \) (common).
  3. Thus, \( \triangle ADE \sim \triangle ABC \) (AAA similarity).
  4. From similarity: \[ \frac{AD}{AB} = \frac{AE}{AC}. \] On simplifying, we again get \[ \frac{AD}{DB} = \frac{AE}{EC}. \]

Hence proved (by similarity).

Converse of Thales’ Theorem

If a line intersects sides \( AB \) and \( AC \) of \( \triangle ABC \) at \( D \) and \( E \) such that \( \dfrac{AD}{DB} = \dfrac{AE}{EC} \), then that line is parallel to \( BC \).

Quick corollaries
  • From similarity: \( \dfrac{AD}{AB} = \dfrac{AE}{AC} = \dfrac{DE}{BC} \).
  • If \( D \) is the midpoint of \( AB \) and \( DE\parallel BC \), then \( E \) is the midpoint of \( AC \).

Worked Example

Example 1: In \( \triangle ABC \), \( DE\parallel BC \). If \( AD=4\,\text{cm} \), \( DB=6\,\text{cm} \) and \( AE=5\,\text{cm} \), find \( EC \).

By BPT: \( \dfrac{AD}{DB} = \dfrac{AE}{EC} \) ⇒ \( \dfrac{4}{6} = \dfrac{5}{EC} \).
⇒ \( EC = \dfrac{6\times5}{4} = 7.5\,\text{cm}. \)

Example 2: In \( \triangle ABC \), \( AB=12\,\text{cm} \), \( AC=15\,\text{cm} \). A point \( D \) on \( AB \) is chosen so that \( AD=8\,\text{cm} \). A line through \( D \) meets \( AC \) at \( E \) and is parallel to \( BC \). Find \( AE \).

From similarity: \( \dfrac{AD}{AB} = \dfrac{AE}{AC} \) ⇒ \( \dfrac{8}{12} = \dfrac{AE}{15} \).
⇒ \( AE = 10\,\text{cm}. \)

Practice Questions

  1. In \( \triangle ABC \), \( DE\parallel BC \). If \( AD:DB = 3:2 \) and \( AE=9\,\text{cm} \), find \( EC \).
  2. In \( \triangle ABC \), \( DE\parallel BC \). If \( AB=18\,\text{cm} \), \( AC=21\,\text{cm} \), and \( AD=12\,\text{cm} \), find \( AE \).
  3. Given \( \triangle ABC \), points \( D\in AB \) and \( E\in AC \) satisfy \( \dfrac{AD}{DB}=\dfrac{AE}{EC}=\dfrac{2}{3} \). Prove \( DE\parallel BC \).

Note: Some books also use the name “Thales’ Theorem” for the circle result: “An angle in a semicircle is a right angle.” In Class 10 NCERT, Thales’ Theorem refers to the Basic Proportionality Theorem.

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