Question

In the figure $DE\parallel AB$ and $DF\parallel AC$. Prove that $EF\parallel BC$ and $BC=EF$

Answer 1

Given:

$AB||DE,AB=DE,AC||DF$ and $AC=DF$

To prove: $BC||EF$ and $BC=EF$

Proof: $AC||DF$ [Given]

And $AC=DF$ [Given]

$\therefore ACFD$ is a parallelogram.

$AD||CF$.......(1) ($\because$ opposite side of a || gm are parallel)

and $AD=CF$.......(2) ($\because$ opposite side of a || gm are parallel)

Now, $AB||DE$ [Given]

and $AB=DE$ [Given]

$\therefore ABED$ is a parallelogram.

$AD||BE$.......(3) ($\because$ opposite side of a || gm are parallel)

and $AD=BE$.........(4) ($\because$ opposite side of a || gm are parallel)

From (1) and (3), we get,

$CF||BE$

From (2) and (4), we get,

$CF=BE$

$\therefore BCFE$ is a parallelogram.

$\Rightarrow BC||EF$ ($\because$ opposite side of a || gm are parallel)

and $BC=EF$ ($\because$ opposite side of a || gm are

hence proved.