In the square $ABCD$ the point $E$ is located on the side $[AB]$, and $F$ is the foot of the perpendicular from $B$ on the line $DE$. The point $L$ belongs to the line $DE$, such that $F$ is between $E$ and $L$, and $FL = BF$. $N$ and $P$ are symmetric of the points $A , F$ with respect to the lines $DE, BL$, respectively. Prove that: a) The quadrilateral $BFLP$ is square and the quadrilateral $ALND$ is rhombus. b) The area of the rhombus $ALND$ is equal to the difference between the areas of the squares $ABCD$ and $BFLP$.