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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 ∥*

*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)*

*hence*

*△*

*ADE*

*≅ △*

*CFE (by SAS)*

*Therefore,*

*∠*

*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 =*

*12*

*BC (DE = EF by construction)*

*Hence proved.*