Prove: the segment joining the midpoints of two sides of a triangle is parallel to the third side.

As per theorem Segment joining midpoints of two sides of a triangle is parallel to the third side.

Step-by-step 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.