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Using converse of B.P.T., prove that the line joining the mid-points of any two sides of a triangle is parallel to the third side.
To do:
Using converse of B.P.T., we have to prove that the line joining the mid-points of any two sides of a triangle is parallel to the third side.
Solution:
Let in a $\triangle ABC$, $D$ be the mid-point of $AB$.
In $\triangle ABC$,
This implies,
$\frac{AD}{DB}=1$.........(i)
$\frac{AE}{EC}=1$........(ii)
Therefore,
$\frac{AD}{DB}=\frac{AE}{EC}$
This implies, by converse of B.P.T.,
$DE \| BC$
Hence proved.
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