prove that a line drawn through the midpoint of one side of a triangle parallel to another cell bisects the third side using basic proportionality theorem
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Given, in ΔABC, 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.
We have to prove that E is the mid point of AC.
Since, D is the mid-point of AB.
∴ AD=DB
⇒AD/DB = 1 …………………. (i)
In ΔABC, DE || BC,
By using Basic Proportionality Theorem,
Therefore, AD/DB = AE/EC
From equation (i), we can write,
⇒ 1 = AE/EC
∴ AE = EC
Hence, proved, E is the midpoint of AC.
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