In ABC ; D and E are mid-points of the sides AB and AC respective… - Vidyadip

Question

In $\triangle ABC ; D$ and $E$ are mid$-$points of the sides $AB$ and $AC$ respectively. Through $E$, a straight line is drawn parallel to $AB$ to meet $BC$ at $F.$Prove that $\text{BDEF}$ is a parallelogram. If $AB = 16 \ cm, AC = 12 \ cm$ and $BC = 18 \ cm$,find the perimeter of the parallelogram $\text{BDEF}.$

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Answer

The figure is shown below

From figure since $E$ is the midpoint of $A C$ and $E F \| A B$
Therefore $F$ is the midpoint of $BC$ and $2 DE = BC$ or $DE = BF$
Again $D$ and $E$ are midpoints,
therefore $DE \| BF$ and $EF = BD$
Hence $\text{BDEF}$ is a parallelogram.
Now,
$BD = EF =\frac{1}{2} AB =\frac{1}{2} \times 16=8 \ cm$
$ BF = DE =\frac{1}{2} BC =\frac{1}{2} \times 18=9 \ cm$
Therefore perimeter of $BDEF =2(B F+E F)=2(9+8)=34 \ cm$.

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