Let’s consider triangle ∆ABC, where D and E are the midpoints of sides AB and AC respectively, and DE is the line segment joining them. We need to prove that DE ∥ BC and DE = \(\frac{1}{2}\) BC. **Proof:** Since D is the midpoint of AB, ∴ AD = DB or, \(\frac{AD}{DB} = 1\) Similarly, since E is the midpoint of AC, ∴ AE = EC or, \(\frac{AE}{EC} = 1\) So, \(\frac{AD}{DB} = \frac{AE}{EC}\) Now, in triangle ∆ABC, since \(\frac{AD}{DB} = \frac{AE}{EC}\), ∴ DE ∥ BC (Proved) Now, in triangles ∆ADE and ∆ABC, ∠ADE = ∠ABC [Because DE ∥ BC and AD is a transversal] ∠AED = ∠ACB [Because DE ∥ BC and AE is a transversal] And ∠DAE = ∠BAC [Common angle] ∴ ∆ADE ∼ ∆ABC (Similar triangles) So, \(\frac{AD}{AB} = \frac{AE}{AC} = \frac{DE}{BC}\) But \(\frac{AD}{AB} = \frac{AE}{AC} = \frac{1}{2}\) [Since D and E are midpoints of AB and AC] ∴ \(\frac{DE}{BC} = \frac{1}{2}\) i.e., DE = \(\frac{1}{2}\) BC (Proved)