State Midpoint Theorem Class 9

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

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