Question 123
Given the square $ABCD$, let $E$ be a point on the side $[AB]$, and $F$ be the foot of the perpendicular line from $B$ on the line $DE$. Let $L$ be a point on the line $DE$ so that $F$ is situated between $E$ and $L$, and $FL = BF$. If $N$ is the symmetric of $A$ with respect to the line $DE$ and $P$ is the symmetric of $F$ with respect to $BL$, prove that:
a) $BFLP$ is a square and $ALND$ is a rhombus;
b) the area of the rhombus $ALND$ is equal with the difference between the areas of the squares $ABCD$ and $BFLP$.
a) $BFLP$ is a square and $ALND$ is a rhombus;
b) the area of the rhombus $ALND$ is equal with the difference between the areas of the squares $ABCD$ and $BFLP$.