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Using B.P.T., prove that a line drawn through the mid-point of one side of a triangle parallel to another side bisects the third side.
To do:
We have to prove that a line drawn through the mid-point of one side of a triangle parallel to another side bisects the third side.
Solution:
We know that,
If a line divides two sides of a triangle proportionally, then it is parallel to the third side.
Let in a $\triangle ABC$, $D$ be the mid-point of $AB$ and $DE \| BC$
In $\triangle ABC, DE \| BC$,
This implies,
$\frac{AD}{DB}=\frac{AE}{EC}$.........(i)
$AD=DB$
This implies,
$\frac{AD}{AD}=\frac{AE}{EC}$
$1=\frac{AE}{EC}$
$AE = EC$
Therefore, $DE$ bisects $AC$
Hence proved.
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