${\text{ABCD}}$ is a parallelogram. The sides ${\text{AB}}$and ${\text{AD}}$are produced to ${\text{E}}$ and ${\text{F}}$respectively, such that ${\text{AB = BE}}$ and ${\text{AD = DF}}$.
Hence, $\Delta {\text{BEC}} \cong \Delta {\text{DCF}}$
If the above statement is true then mention the answer as $1$,else mention $0$ if false.

Hint:-In this question you first need to make relations between different angles, sides and given condition to get the relation between the triangles $\Delta {\text{BEC and }}\Delta {\text{DCF}}$and then checking the condition that these are congruent or not, using SAS(side-Angle-Side), SSS(side-side-side), ASA(Angle-Side-Angle) properties.

Complete step-by-step answer:

Given: ABCD is a parallelogram.
And AB=BE , AD=DF
We know,
AD=BC ($\therefore $ABCD is a parallelogram)
And AD=DF ($\therefore $Given)
Thus,
DF=BC eq.1
Similarly
AB=CD ($\therefore $ABCD is a parallelogram)
And AB=BE ($\therefore $Given)
Thus,
BE=CD eq.2
Since, corresponding angles for parallel lines of a parallelogram are equal
Then,

$ \Rightarrow \angle {\text{BAD = }}\angle {\text{CDF (}}\therefore {\text{corresponding angles for parallel lines AB and CD)}} \\
   \Rightarrow \angle {\text{BAD = }}\angle {\text{CBE (}}\therefore {\text{corresponding angles for parallel lines AD and BC)}} \\ $

Thus,
\[\angle {\text{CDF = }}\angle {\text{CBE eq}}{\text{.3}}\]
Now, consider \[\Delta {\text{DCF and }}\Delta {\text{BEC}}\]
DF=BC {$\therefore $from eq.1}
\[\angle {\text{CDF = }}\angle {\text{CBE}}\] {$\therefore $from eq.3}
CD=BE {$\therefore $from eq.2}
Thus, \[{\text{ }}\Delta {\text{DCF }} \cong {\text{ }}\Delta {\text{BEC}}\] {$\therefore $Side-Angle-Side Rule}
Since, the given statement is true then the answer is 1.

Note:-Whenever you get this type of problem the key concept of solving is to make a rough diagram of the situation which is given . Then using properties of given shape (as in this question parallelogram) and SSS,SAS, ASA etc. form relations between different angles and sides to get the required result.