Variant of Japanese theorem for a cyclic quadrilateral

Question

Let $ABCD$ be a convex cyclic quadrilateral, let $A', B', C', D'$ lie on angle bisector (all internal or all external) of $DAB$, $ABC$, $BCD$, $CDA$. Let $AA'B'B$ , $BB'C'C$, $CC'D'D$, $DD'A'A$ are cyclic quadrilateral then show that: $A'B'C'D'$ is a rectangle.

Answer 1

Since $ABA^\prime B^\prime$ is cyclic $$\angle AA^\prime B^\prime=\angle AB B^\prime=\angle B/2.$$ Similarly, since $ADA^\prime D^\prime$ is cyclic, $$\angle AA^\prime D^\prime=\angle ADD^\prime=\angle D/2$$ it follows that $$\angle D^\prime A^\prime B^\prime=\angle B/2+\angle D/2=90^\circ.$$ Similarly for the other angles.