Question

If a line drawn parallel to one side of a triangle intersecting the other two sides in distinct points divides the two sides in the same ratio, then it is parallel to the third side. State and prove the converse of the above statement.

Hint:

To prove parallel lines using proportional sides, apply the Converse of the Basic Proportionality Theorem.

Answer 1

Statement (Converse):
If a line divides two sides of a triangle in the same ratio, then it is parallel to the third side.

Given:
In \(\triangle ABC\), a line intersects \(AB\) at \(D\) and \(AC\) at \(E\) such that
\[ \frac{AD}{DB} = \frac{AE}{EC} \] We need to prove that \(DE \parallel BC\).

Proof:
Assume \(DE\) is not parallel to \(BC\).
Draw a line \(D' E'\) through \(D\) parallel to \(BC\), intersecting \(AC\) at \(E'\).

By Basic Proportionality Theorem (Thales theorem):
\[ \frac{AD}{DB} = \frac{AE'}{E'C} \] But from given,
\[ \frac{AD}{DB} = \frac{AE}{EC} \] Thus,
\[ \frac{AE'}{E'C} = \frac{AE}{EC} \] Since \(E'\) and \(E\) divide \(AC\) in the same ratio, points \(E\) and \(E'\) coincide.
Hence, \(E = E'\).

Therefore, the line through \(D\) and \(E\) is parallel to \(BC\).

Conclusion:
If a line divides two sides of a triangle in the same ratio, then it is parallel to the third side.
Hence, the converse of the Basic Proportionality Theorem is proved.