*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.*