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 $$OP^4 + \left(\frac{MN}{2} \right)^4 = MP^2 \cdot NP^2$$
Source: 2011 Saudi Arabia BMO TST 3.1 - Balkan MO
Tags: geometry, square
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 $$OP^4 + \left(\frac{MN}{2} \right)^4 = MP^2 \cdot NP^2$$