If the midpoints of any sides of a triangle are joined by a line segment, then this line segment will be parallel to the remaining side and will measure half of the remaining side.
Let us consider ABC is a triangle as shown in the following figure:
Let D and E be the midpoints of AB and AC. Then line DE is parallel to BC and DE is half of BC; i.e.
DE ∥ BC
DE = 1/2 BC.
The proof of mid point theorem is as follows.
Have a look at the following diagram:
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 = 12 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)
△ ADE ≅ △ CFE (by SAS)
∠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 = 12BC (DE = EF by construction)