Here, In △△ ABC, D and E are the midpoints of sides AB and AC respectively. D and E are joined. Given: AD = DB and AE = EC. To Prove: DE ∥∥ BC and DE = 1212 BC. Construction: Extend line segment DE to F such that DE = EF. Proof: In △△ ADE and △△ CFE AE = EC (given) ∠∠AED = ∠∠CEF (vertically opposite angles) DE = EF (construction) hence △△ ADE ≅≅ △△ CFE (by SAS) Therefore, ∠∠ADE = ∠∠CFE (by c.p.c.t.) ∠∠DAE = ∠∠FCE (by c.p.c.t.) and AD = CF (by c.p.c.t.) The angles ∠∠ADE and ∠∠CFE are alternate interior angles assuming AB and CF are two lines intersected by transversal DF. Similarly, ∠∠DAE and ∠∠FCE are alternate interior angles assuming AB and CF are two lines intersected by transversal AC. Therefore, AB ∥∥ CF So, BD ∥∥ CF and BD = CF (since AD = BD and it is proved above that AD = CF) Thus, BDFC is a parallelogram. By the properties of parallelogram, we have DF ∥∥ BC and DF = BC DE ∥∥ BC and DE = 1212BC (DE = EF by construction) Hence proved.