Question

Prove that
The line drawn through the mid point of one side of a triangle ,parallel to another side bisects the third side?    

Answer 1

Let in triangle ABC, D is the mid point of AB and DE is parallel to BC.
Now to prove: AE=EC

Construct CF parallel To BA that meet DE produced at F.
Now , DE is parallel to BC (Given)

BD is parallel to CF (by construction)

So, BCFD is a parallelogram.(both  pair of opposite sides are equal)

∴ BD=CF  --------->(1)          (opposite sides of parallelogram)

and BD= AD -------(2)           (D is the mid point of AB)
 
From equation (1) and (2) we get,
AD=CF

Now in ΔADE and Δ CFE,
AD = CF,

∠ADE =∠ CFE    (alternate angles),

∠AED =∠CEF      (vertically opposite angles),

∴ Δ ADE is congurent to ΔCFE  (by AAS)

so ,AE=EC,

∴ DE bisects AC.

Answer 2

Given,In triangle ABC, D is the midpoint of AB such that AD=DB.

A line parallel to BC intersects AC at E as shown in above figure such that DE||BC.

To prove, E is the midpoint of AC.

Since, D is the midpoint of AB

So,AD=DB

⇒ AD/DB=1.....................(i)

In triangle ABC,DE||BC,

By using basic proportionality theorem,

Therefore, AD/DB=AE/EC

From equation 1,we can write,

⇒ 1=AE/EC

So,AE=EC

Hence, proved,E is the midpoint of AC.