Question

BE = DF

If the above statement is true then mention answer as 1, else mention 0 if false

Answer 1

Given, $AB\parallel FD$ and $AC\parallel GE$
$BD=CE$
$BD+DE=CE+DE$
$BE=CD$..(I)
In $△BGE$ and $△CFD$,
$BE=CD$ (From I)
$\angle GBE=\angle FDC$ (Corresponding angles)
$\angle GEB=\angle FCD$ (Corresponding angles)
Thus, $△BGE\cong △DFC$ (SAS rule)
Hence, $BG=DF$ (By cpct)