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State the converse of "Basic Proportionality Theorem" and use it to prove the following :
Line segment joining mid-points of any two sides of a triangle is parallel to the third side.
Line segment joining mid-points of any two sides of a triangle is parallel to the third side.
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Solution: Statement : If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.
Given : In $\Delta ABC$, $D$ and $E$ are mid-points of $AB$ and $AC$ respectively
To prove : $DE \parallel BC$
Proof : As $D$ is the mid-point of $AB$
$AD = DB \Rightarrow \frac{AD}{DB} = 1$
Similarly, $E$ is the mid-point of $AC$
$\Rightarrow \frac{AE}{EC} = 1$
$\therefore \frac{AD}{DB} = \frac{AE}{EC}$
By converse of BPT, $DE \parallel BC$
Given : In $\Delta ABC$, $D$ and $E$ are mid-points of $AB$ and $AC$ respectively
To prove : $DE \parallel BC$
Proof : As $D$ is the mid-point of $AB$
$AD = DB \Rightarrow \frac{AD}{DB} = 1$
Similarly, $E$ is the mid-point of $AC$
$\Rightarrow \frac{AE}{EC} = 1$
$\therefore \frac{AD}{DB} = \frac{AE}{EC}$
By converse of BPT, $DE \parallel BC$