Question

Question 7
Using Basic proportionality theorem, prove that a line drawn through the mid-points of one side of a triangle parallel to another side bisects the third side.

Answer 1

Given, ΔABC in which D is the mid-point 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 that E is the mid-point of AC.
Proof
D is the mid-point of AB.
$\therefore AD=DB$
$\Rightarrow \frac{AD}{BD}=1...\left(i\right)$
In $\Delta ABC,DE||BC$,
Therefore, $\frac{AD}{DB}=\frac{AE}{EC}$ [By using Basic Proportionality Theorem]
$\Rightarrow 1=\frac{AE}{EC}$ [From equation (i)]
$\therefore AE=EC$
Hence, E is the mid-point of AC.