Letting $D,E,F$ be the projections of $I$ on the sidelegths and $M_A,M_B,M_C$ those of $O$, we note that $O$ being the midpoint of $IP$ implies $M_A$ being the midpoint of $DX$, or in other words $CX=BD=\frac{a+c-b}{2}$ and similarly for the other segments. Therefore, by Pythagoras, $$BP^2+PY^2-CP^2-PZ^2=(BX^2+XP^2)-(CX^2+XP^2)+(AP^2-AY^2)-(AP^2-AZ^2)=BX^2-CX^2+AZ^2-AY^2$$$$=CD^2-BD^2+BF^2-CE^2=0$$since $BD=BF$ and $CE=CD$. Therefore all three equalities cyclically hold and we're done.