Prove that a line draw through the mid point of one side of a triangle parallel to another side bisects the third side .

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Solution

Given : A $△$ ABC in which D is the mid-point of side AB and the line DE is drawn parallel to BC, meeting AC in E.
To prove : E is the mid-point of AC i.e., AE = EC.
Proof :
In $△$ ABC, we have,
$D E ∥ B C$
$⟹ \frac{A D}{D B} = \frac{A E}{E C}$ [By Thale's Theorem] ...(1)

But, D is the mid-point of AB.
$⟹ A D = D B$
$\frac{A D}{D B} = 1$ ...(2)
From equation 1 & 2, we get,
$\frac{A E}{E C} = 1$
$A E = E C$
Hence, $E$ bisects $A C$ .

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