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 $1$: Consider quadrilateral $ABED$. Since $AB = DE$ and $AB \parallel DE

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(N/A) Given: $AB = DE$,$AB \parallel DE$,$BC = EF$,and $BC \parallel EF$.
Step $1$: Consider quadrilateral $ABED$. Since $AB = DE$ and $AB \parallel DE$,one pair of opposite sides is equal and parallel. Therefore,$ABED$ is a parallelogram.
Step $2$: Since $ABED$ is a parallelogram,$AD = BE$ and $AD \parallel BE$.
Step $3$: Consider quadrilateral $BCFE$. Since $BC = EF$ and $BC \parallel EF$,one pair of opposite sides is equal and parallel. Therefore,$BCFE$ is a parallelogram.
Step $4$: Since $BCFE$ is a parallelogram,$BE = CF$ and $BE \parallel CF$.
Step $5$: From Step $2$ and Step $4$,we have $AD = BE$ and $BE = CF$,which implies $AD = CF$. Also,$AD \parallel BE$ and $BE \parallel CF$,which implies $AD \parallel CF$.
Step $6$: In quadrilateral $ACFD$,since $AD = CF$ and $AD \parallel CF$,it is a parallelogram.
Step $7$: Since $ACFD$ is a parallelogram,its opposite sides are equal. Therefore,$AC = DF$.

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 $AC = DF$.

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