Question

proved mid point theorem
​

Answer 1

Mid point Theorem :

The line segment joining the mid points of any two sides of a triangle is parallel to the third side.

Given :

A \triangle ABC△ABC in which D and E are the mid points of AB and AC, respectively.

To prove :

DE \parallel BCDE∥BC.

Proof :

Since D and E are the mid points of AB and AC, respectively, we have AD=DBAD=DB and AE=ECAE=EC.

Therefore,

\dfrac{AD}{DB}=\dfrac{AE}{EC}  

DB

AD

​  

=  

EC

AE

​  

           ( each equal to 1 )

Therefore, by the converse of thales theorem, DE \parallel BCDE∥BC.

Answer 2

Construction- Extend the line segment DE and produce it to F such that, EF=DE.

In the triangle, ADE, and also the triangle CFE

EC= AE —– (given)

∠CEF = ∠AED {vertically opposite angles}

EF = DE { by construction}

hence,

△ CFE ≅ △ ADE {by SAS}

Therefore,

∠CFE = ∠ADE {by c.p.c.t.}

∠FCE= ∠DAE {by c.p.c.t.}

and CF = AD {by c.p.c.t.}

The angles, ∠CFE and ∠ADE are the alternate interior angles. Assume CF and AB as two lines which are intersected by the transversal DF.

In a similar way, ∠FCE and ∠DAE are the alternate interior angles. Assume CF and AB are the two lines which are intersected by the transversal AC.

Therefore, CF ∥ AB

So, CF ∥ BD

and CF = BD {since BD = AD, it is proved that CF = AD}

Thus, BDFC forms a parallelogram.

By the use of properties of a parallelogram, we can write

BC ∥ DF

and BC = DF

BC ∥ DE

and DE = (1/2 * BC).

Hence, the midpoint theorem is Proved.