Question

Using bpt prove that a line drawn through the mid point of one side of a triangle parallel to another side bisects the third side

Answer 1

Given:A ΔABC,in which D is the midpoint of AB and DE║BC
To prove:DE=1/2 BC
Proof:In ΔADE and ΔABC,
∠A=∠A   (common)
∠ADE=∠ABC  (corres.∠s as DE║BC)
∴ΔADE~ΔABC  (by AA similarity)
∴AD/DE=AB/BC   (By BPT)
⇒AD/DE=2 AD/BC    (D is the midpoint of AB⇒AD=(1/2)AB⇒AB=2 AD)
AD×BC=2 AD×DE⇒(AD×BC)-(2 AD×DE)=0
⇒AD(BC-2 DE)=0⇒BC-2 DE=0⇒BC=2 DE⇒DE=1/2 BC.......thus,proved

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.