Question

prove mid point theorem

Answer 1

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 = 1212 BC.

Construction: Extend line se
gment 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 = 1212BC  (DE = EF by construction)

Hence proved.

Answer 2

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.