Answer:
As per theorem Segment joining midpoints of two sides of a triangle is parallel to the third side.
Explanation:
Given : In a triangle ABC, D and E are the midpoint of sides AB and AC.
To Proof : DE║BC
Construction: A line CF parallel to BA is drawn and De is extended to point F.
Proof : In a triangle ADE and EFC
Sides AD║CF and DF is transverse.Then ∠ADE = ∠CFE (Alternate angles)
Similarly AD║CF and AC is transverse then ∠DAE = ∠FCA (Alternate angles)
Therefore ΔADE≅ΔCEF (AAS property)
From this property side AD = Side CF (corresponding sides of the congruent triangles)
Therefore BD = CF
So sides BD and CF are equal and parallel to each other which represents DBCF is a parallelogram.
Therefore DE║BC
Hence proved.