Theorem – The line segment joining the mid-points of two sides of a triangle is parallel to the third side.

Given: A triangle △ABC, in which D and E are mid points of sides AB and AC respectively. DE is joined.

To prove: Line joining the mid points D and E (DE) is parallel to the third line BC. 

                             DE ∥ BC   

Construction: Through D, draw a line segment parallel BC to meet DE to F, such that DE = EF.  Join FC.

Proof: In △ AED and △ CEF, 

AE = CE (E is the mid point of AC)

∠AED = ∠CEF (vertically opposite angle)

DE = EF (By construction)

∴ △ AED ≅ △ CFE (SAS rule)

∴ AD = FC …..(1) (c.p.c.t.)

∠ADE = ∠CFE…..(2) (c.p.c.t.)

Now, D is the mid point of AB.

 AD = DB ….(3)

From (1) and (3)

FC = DB…..(4)

From (2)    

∠ADE = ∠CFE (pair of alternate interior angles are equal then lines are parallel) 

∴ AD ∥ FC ….(5)

 From (1) and (5)

AD = FC and AD ∥ FC ….(6)

From (6) we find that DBCF is a parallelogram, because one pair of opposite sides are equal and parallel.

∴ DBCF is a parallelogram. 

∴ BC = DF and BC ∥ DF,

  ∴ BC ∥ DE  

Line joining the mid points D and E is parallel to the third line BC. 

                             DE ∥ BC

  Hence Proved.