Question

In ${△}^{s}ABC$ and $DEF,AB\parallel DE;BC=EF$ and $BC\parallel EF$. Vertices, $A,B$ and $C$ are joined to vertices $D,E$ and $F$ respectively (see figure). Show that
$△ABC\cong △DEF$

Answer 1

Given: $\Delta ABC$ and $\Delta DEF,AB\left|\right|DE;AB=DE;BC=EF;BC\left|\right|EF$

To show: $\Delta ABC\cong \Delta DEF$

Proof:

As we know that $ABCD$ is a parallelogram, so, $AD||BE$ and $AD=BE$

As we know that $BEFC$ is a parallelogram, so, $BE\left|\right|CF$ and $BE=CE.$

Now,if $AD=BE$ and $BE=CF$ so, $AD||CF$

Also, if $AD=BE$ and $BE=CF$ so, $AD=CF$.

This, implies one pair of opposite sides are equal and parallel to each other.

Therefore, $ADFC$ is a parallelogram.

Therefore, $AC=DF$

In triangles $ABC$ and $DEF$,

$AB=DE$ (opposites sides of parallelogram)

$BC=EF$ (opposites sides of parallelogram)

$AC=DF$ (proven above)

So, by SSS criterion of congruency,triangle $ABC$ is congruent to triangle $DEF$.