In Delta ABC and Delta DEF,AB = DE,AB paralle | vedclass

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(N/A) Given: $AB = DE$,$AB \parallel DE$,$BC = EF$,and $BC \parallel EF$.Step $1$: Since $AB = DE$ and $AB \parallel DE$,quadrilateral $ABED$ is a par

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(N/A) Given: $AB = DE$,$AB \parallel DE$,$BC = EF$,and $BC \parallel EF$.
Step $1$: Since $AB = DE$ and $AB \parallel DE$,quadrilateral $ABED$ is a parallelogram. Therefore,$AD = BE$ and $AD \parallel BE$.
Step $2$: Since $BC = EF$ and $BC \parallel EF$,quadrilateral $BEFC$ is a parallelogram. Therefore,$BE = CF$ and $BE \parallel CF$.
Step $3$: From Step $1$ and Step $2$,we have $AD = BE$ and $BE = CF$,which implies $AD = CF$. Also,$AD \parallel CF$ because both are parallel to $BE$.
Step $4$: Since $AD = CF$ and $AD \parallel CF$,quadrilateral $ACFD$ is a parallelogram. Therefore,$AC = DF$.
Step $5$: In $\Delta ABC$ and $\Delta DEF$:
$AB = DE$ (Given)
$BC = EF$ (Given)
$AC = DF$ (Proved in Step $4$)
Therefore,by $SSS$ congruence criterion,$\Delta ABC \cong \Delta DEF$.

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

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