Point $X$ is missed, I think it is point of intersection of circumcircles of triangles $B_1A_1C$ and $A_1D_1B$ Let $AB=c,BC=a,CA=b,a^2+b^2=c^2$ Not hard to prove, that $A_1B_1=\frac{abc}{a^2+b^2+ab}$ Let $B_1Y$ meet hypotenuse $AB$ at $R$ $\angle B_1YA_1=\angle B_1YA = 90^o=$ so $Y$ lies on $AA_1$ and $\angle C_1B_1R=\angle B_1A_1A=\angle A_1AR$ $AC_1=B_1C_1 \cot \angle BAC=\frac{b^2c}{a^2+b^2+ab}$ $C_1R=B_1C_1 \tan \angle C_1B_1R = \frac{A_1B_1^2}{A_1B_1+AC_1}=\frac{A_1B_1^2}{A_1B_1+ A_1B_1\frac{AC}{BC}}=\frac{A_1B_1BC}{BC+AC}=\frac{abc}{a^2+b^2+ab} \frac{a}{a+b}=\frac{a^2bc}{(a+b)(a^2+b^2+ab)}$ $AR=AC_1+C_1R=\frac{b^2c}{a^2+b^2+ab}+\frac{a^2bc}{(a+b)(a^2+b^2+ab)} = \frac{bc}{a^2+b^2+ab} ( b+\frac{a^2}{a+b})=\frac{bc}{a+b}$ Let $A_1X$ and $AB$ meets at $Q$. Same way we can prove $QB=\frac{ac}{a+b}$ $AR+QB=c=AB \to R$ and $Q$ are same point ( which is also point of intersection of angle bisector of $\angle ACB$ with $AB$)

Let J be center of square. We have some simple collinearities. A,Y,A1 and A,J,X and B,X,B1 and B,J,Y are all collinear. Let B1Y meet C1D1 at Z. we need to prove B1C1ZX is cyclic but first we will prove that AC1XA1 is cyclic. ∠AC1X = 135 = ∠JXA1 = ∠AXA1 ---> AC1XA1 is cyclic. ∠XB1Z = ∠XA1Y = ∠XC1Z ---> B1C1ZX is cyclic ---> ZXB1 = 90 ---> Z,X,A1 are collinear so we're Done.