Question

State and prove Basic Proportionality Theorem.

Hint:

Remember: In Basic Proportionality Theorem, a line parallel to one side of a triangle divides the other two sides proportionally.

Answer 1

Step 1: State the Basic Proportionality Theorem.}
The Basic Proportionality Theorem states that if a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, then the other two sides are divided in the same ratio.

Step 2: Write the given figure and assumption.}
In \( \triangle ABC \), let a line \( DE \) be drawn parallel to side \( BC \), such that \( D \) lies on \( AB \) and \( E \) lies on \( AC \). Then we have to prove that \[ \frac{AD}{DB} = \frac{AE}{EC} \]
Step 3: Compare the areas of triangles with the same altitude.}
Since triangles \( \triangle BDE \) and \( \triangle CDE \) lie on the same base \( DE \) and between the same parallels \( DE \parallel BC \), their areas are equal. Therefore, \[ \text{ar}(\triangle BDE) = \text{ar}(\triangle CDE) \] Also, triangles \( \triangle ADE \) and \( \triangle BDE \) have the same altitude from \( E \) on line \( AB \). Hence, \[ \frac{\text{ar}(\triangle ADE)}{\text{ar}(\triangle BDE)} = \frac{AD}{DB} \] Similarly, triangles \( \triangle ADE \) and \( \triangle CDE \) have the same altitude from \( D \) on line \( AC \). Hence, \[ \frac{\text{ar}(\triangle ADE)}{\text{ar}(\triangle CDE)} = \frac{AE}{EC} \]
Step 4: Use the equality of areas.}
Since \[ \text{ar}(\triangle BDE) = \text{ar}(\triangle CDE) \] therefore, \[ \frac{\text{ar}(\triangle ADE)}{\text{ar}(\triangle BDE)} = \frac{\text{ar}(\triangle ADE)}{\text{ar}(\triangle CDE)} \] So, we get \[ \frac{AD}{DB} = \frac{AE}{EC} \]
Step 5: Write the conclusion.}
Hence, it is proved that if a line is drawn parallel to one side of a triangle, then it divides the other two sides in the same ratio. Therefore, \[ \boxed{\frac{AD}{DB} = \frac{AE}{EC}} \]