Denote $x = \frac{AX}{AB},\ y = \frac{AY}{AC},\ z = \frac{AZ}{AD}$. AY bisects the right angle $\angle XAZ$, hence $AY = \frac{AX \cdot AZ\ \sqrt 2}{AX + AZ}$ and $y = \frac{AY}{AB \sqrt 2} = \frac{xz}{x + z}.$ (O) is the excircle of the right angle triangle $\triangle XAZ.$ Therefore, $AX + XY + AZ = 2AE = AB$ and $XY = \sqrt{AX^2 + AZ^2}$, which implies $x + \sqrt{x^2 + z^2} + z = 1$ $x^2 + z^2 = (1 - x - z)^2 = 1 + x^2 + z^2 - 2z - 2x + 2xz$ $x + z - xz = \frac 1 2,\ \ \ (1 - x)(1 - z) = 1 - x - z + xz = \frac 1 2$ Substituting the above into the sum $\frac{AX}{XB} + \frac{AY}{YC} + \frac{AZ}{ZD},$ $\frac{x}{1 - x} + \frac{y}{1 - y} + \frac{z}{1 - z} = \frac{x}{1 - x} + \frac{xz}{x + z - xz} + \frac{z}{1 - z} =$ $= \frac{x}{1 - x} + 2xz + \frac{z}{1 - z} = \frac{x(1 - z) + 2xz(1 - x)(1 - z) + z(1 - x)}{(1 - x)(1 - z)} =$ $= \frac{x - xz + xz + z - xz}{\frac 1 2} = 2(x + z - xz) = 1$