Question

If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, then prove that the other two sides are divided in the same ratio.

Answer 1

Step 1: Understanding the problem:
We are given a triangle $△ABC$ with a line drawn parallel to one of its sides, say $BC$. This line intersects the other two sides, $AB$ and $AC$, at distinct points, say $P$ and $Q$ respectively. We are asked to prove that the segments $AP$ and $PB$ on side $AB$, and $AQ$ and $QC$ on side $AC$, are in the same ratio.

Step 2: Applying Basic Proportionality Theorem:
The Basic Proportionality Theorem (also called 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, then the other two sides are divided in the same ratio.
In our case, since line $PQ$ is parallel to side $BC$ of triangle $△ABC$, we apply this theorem:

\[ \frac{AP}{PB} = \frac{AQ}{QC} \] This shows that the segments on the two sides $AB$ and $AC$ are divided in the same ratio by the parallel line $PQ$.

Step 3: Conclusion:
By the Basic Proportionality Theorem, we have proven that the other two sides of the triangle, $AB$ and $AC$, are divided in the same ratio by the line drawn parallel to one side of the triangle.