Question

In $△ABC$ and $△DEF,AB=DE,AB\parallel DE,BC=EF$ and $BC\parallel EF$. Vertices $A,B$ and $C$ are joined to vertices $D,E$ and $F$ respectively. Show that

(i) quadrilateral $ABED$ is a parallelogram
(ii) quadrilateral $BEFC$ is a parallelogram
(iii) $AD\parallel CF$ and $AD=CF$
(iv) quadrilateral $ACFD$ is a parallelogram
(v) $AC=DF$
(vi) $△ABC=△DEF$.

Answer 1

$\left(i\right)$Suppose $BD$ is a diagonal of quadrilateral $ABCD$.

As, $AB=DE\therefore \angle ABD=\angle EDB$(Alternate angles)

In $\Delta ABD$ & $\Delta EDB$

$AB=ED$(Given)

$\angle ABD=\angle EDB$

$BD=BD$(Common)

$\therefore \Delta ABD\cong \Delta EBD$ (SAS rule)

Hence, $BE=AD$ (CPCT)

Thus, in $ABED$ Both pairs of opposite sides are equal

$\therefore ABED$ is a parallelogram.

$\left(ii\right)$Again consider a diagonal $BF$ of quadrilateral $BCFE$.

Similar to (i) we know that $\Delta BCE\cong \Delta EFC$ (SAS rule)

Hence, $BE=CF$ (CPCT)

Thus, in $BCFE$ Both pairs of opposite sides are equal

$\therefore BCFE$ is a parallelogram.

$\left(iii\right)$ From part$\left(i\right)$, we proved that ABED is a parallelogram

So, $AD=CF$ and $AD||CF$

From part$\left(ii\right)$, we proved that BEFC is a parallelogram

So, $BE=CF$ and $BE||CF$

hence from (i) and (ii), $AD||CF$ $AD=CF$

$\left(iv\right)$Again consider a diagonal $AF$ of quadrilateral $ADFC$.

Similar to (i) we know that $\Delta ACD\cong \Delta DFC$ (SAS rule)

Hence, $AC=DF$ (CPCT)

Thus, in $ADFC$ Both pairs of opposite sides are equal

$\therefore ADFC$ is a parallelogram.

As $ADFC$ is a parallelogram $\therefore$ opposite sides are parallel also.

$\left(v\right)$Proved in $\left(iv\right)$

$\left(vi\right)$By above all, we know that $AB=DE,BC=EF,AC=DF$

$\therefore \Delta ABC=\Delta DEF$........(SSS rule)