Question

Proof that a line drawn through the
mid point of one side of a triangle parallel to another side bisect third side

Answer 1

Please draw a triangle as given below

Let us assume triangle ABC
Where DE is parallel to BC and D is the mid-point of AB

To prove: E is the mid-point of AC

Proof:
In ΔABC, DE||BC

We know that if a line drawn parallel to one side of triangle, intersects the other two sides  in distinct points, then it divides the other 2 sides in same ratio

$\frac{AD}{DB} = \frac{AE}{EC}$

$\frac{DB}{DB} = \frac{AE}{EC}$

$1 = \frac{AE}{EC}$

EC = AE 
So E is the mid-point of AC

Hence proved!

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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.