## Midpoint Theorem Statement:

“**The line segment in a triangle joining the midpoint of two sides of the triangle is said to be parallel to its third side and is also half of the length of the third side**“.

**Given: **A △ABC in which D and E are the mid-points of sides AB and AC respectively. DE is joined to F.

**To Prove:** DE || BC and DE = 1/2 BC

**Construction:** Produce the line segment DE to F, such that DE = EF and Join FC.

**Proof: **In △s AED and CEF, we have

⇒ AE = CE [E is the midpoint of AC]

⇒ ∠AED = ∠CEF [Vertically opposite angles]

and, DE = EF [by construction]

So, by SAS criterion of congruence, we have

AED **⩭** ACEF

⇒ AD = CF …………………….(a)

and, ∠ADE = ∠CFE [c.p.c.t.] ………………. (b)

Now, D is the midpoint of AB

⇒ AD = DB

⇒ DB = CF [From (i) AD= CF] ………………. (c)

⇒ ∠ADE = ∠CFE

i.e., alternate interior angles are equal

AD∥FC

DB∥FC ………………. (d)

From eq. (c) and (d), we find that DBCF is a quadrilateral, in which

DF∥BC and DF = BC

But, D, E, F are collinear and DE=EF

∴ DF∥BC and DE = 1/2 BC