Since $AB_1$ and $B_1A_1$ are the diameters of the circumcircles of triangles $AC_1B_1$ and $A_1CB_1$ respectively, $\angle AYB_1 = \angle A_1YB_1 = 90^\circ$. Hence $Y$ belongs to the line $AA_1$ and $AA_1 \perp B_1Y$. Similarly, $X$ belongs to the line $BB_1$ and $BB_1 \perp A_1X$.

Let $Z_b$ and $Z_a$ be the points of intersection of the lines $B_1Y$ and $A_1X$ respectively with the hypotenuse $AB$. Let $CB = a$, $AC = b$, $AB = c$ and $A_1B_1 = x$. We will find the lengths of the segments $BZ_a$ and $AZ_b$, and then prove the equality $AZ_b + Z_aB = AB$ from which follows the statement of the problem.
Since $\angle B_1XZ_a = \angle B_1C_1Z_a = 90^\circ$, the quadrilateral $B_1C_1Z_aX$ is cyclic. The quadrilateral $CA_1XB_1$ is cyclic too, therefore
$$BA_1 \cdot CB = BX \cdot BB_1 = BZ_a \cdot BC_1. \quad (*)$$
The triangle $A_1BD_1$ is similar to the triangle $ABC$, therefore
$$\frac{A_1B}{AB} = \frac{D_1B}{CB} = \frac{A_1D_1}{AC}.$$
Hence $A_1B = \frac{cx}{b}$, $D_1B = \frac{ax}{b}$ and $BC_1 = x + \frac{ax}{b} = \frac{(a+b)x}{b}$. From $(*)$ it follows that $BZ_a = \frac{ac}{a+b}$. Similarly one can prove $AZ_b = \frac{bc}{a+b}$, so $AZ_b + BZ_a = AB$.