Question 80
Let $ABCD$ be a square of center $O$. The parallel to $AD$ through $O$ intersects $AB$ and $CD$ at $M$ and $N$ and a parallel to $AB$ intersects diagonal $AC$ at $P$. Prove that
$$
OP^{4} + \left(\frac{MN}{2}\right)^{4} = MP^{2} \cdot NP^{2}
$$
$$
OP^{4} + \left(\frac{MN}{2}\right)^{4} = MP^{2} \cdot NP^{2}
$$