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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 = 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


and DF = BC


and DE = 12BC  (DE = EF by construction)

Hence proved.
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