Step 1: <b>Converse of Basic Proportionality Theorem (BPT):</b> If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.
Step 2: <b>Given:</b> In $\triangle ABC$, $D$ is the mid-point of $AB$ and $E$ is the mid-point of $AC$.
Step 3: <b>To prove:</b> $DE \parallel BC$.
Step 4: Since $D$ is the mid-point of $AB$,
$$AD=DB \;\Rightarrow\; \frac{AD}{DB}=1$$
Step 5: Since $E$ is the mid-point of $AC$,
$$AE=EC \;\Rightarrow\; \frac{AE}{EC}=1$$
Step 6: From Steps 4 and 5,
$$\frac{AD}{DB}=\frac{AE}{EC}$$
Step 7: So the line $DE$ divides the two sides $AB$ and $AC$ of $\triangle ABC$ in the same ratio.
Step 8: By the converse of the Basic Proportionality Theorem,
$$DE \parallel BC$$
Hence the line segment joining the mid-points of any two sides of a triangle is parallel to the third side. Proved.