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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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