Question

Prove that a line drawn through the mid-point of one side of a triangle parallel to another side bisects the third side (Using basic proportionality theorem).
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Answer 1

Answer:

In △MNS, line KL ∥ side NS     .....(given)

∴KNMK=LSML      ..... (By BPT)

But MK=KN

∴KNMK=1

∴LSML=1

∴ML=LS

This means that line KL bisects side MS. 



Answer 2

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.