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.
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: E is the mid point of AC.
Proof: D is the mid-point of AB.
∴ AD=DB
⇒ AD/BD = 1 ... (i)
In ΔABC, DE || BC,
Therefore, AD/DB = AE/EC [By using Basic Proportionality Theorem]
⇒1 = AE/EC [From equation (i)]
∴ AE =EC
Hence, E is the mid point of AC.
