Let $P$ be the point of intersection of the diagonals; let $A',B',C',D'$ be the four reflections across the diagonals. Also let $\ell$ and $m$ be the lines $AC$ and $BD$ and let $\ell'$ and $m'$ the reflection of the two diagonals about the other. For point a, since $P,A',C'\in \ell'$ and $P,B',D'\in m'$, we just need $\ell'\equiv m'$. Therefore, by reflection of lines, $$\measuredangle (\ell,m)=\measuredangle (\ell,m')+\measuredangle (\ell',m)=\measuredangle (m,\ell)+\measuredangle (m,\ell)=-2\measuredangle (\ell,m)$$which implies $3\measuredangle (\ell,m)=0\pmod{\pi}$, which means that the two angles between the diagonals are $\frac{\pi}{3}$ and $\frac{2\pi}{3}$. For point b, by power of point and summery we have $$PA'\cdot PC'=PA\cdot PC=PB\cdot PD=PB'\cdot PD'$$and so $A'B'C'D'$ is cyclic.