, 2019

Solution:
Let $X'$ and $Y'$ lie on segments $AB$ and $AD$ respectively such that $ZX' \parallel AD$ and $ZY' \parallel AB$. We note that triangles $AXY$ and $Y'YZ$ are similar, and that triangles $AY'Z$ and $ADC$ are similar. Thus, we have
$$\frac{AC}{AZ} = \frac{AD}{AY'} \text{ and } \frac{AY'}{AY} = \frac{XZ}{XY}$$
This means that
$$\frac{AD}{AY} = \frac{AD}{AY'} \cdot \frac{AY'}{AY} = \frac{XZ}{XY} \cdot \frac{AC}{AZ}$$
and similarly,
$$\frac{AB}{AX} = \frac{ZY}{XY} \cdot \frac{AC}{AZ}$$
Therefore we have
$$\frac{AB}{AX} + \frac{AD}{AY} = \left(\frac{XZ}{XY} + \frac{ZY}{XY}\right) \cdot \frac{AC}{AZ} = \frac{AC}{AZ}$$
as desired.

Solution:
We recall that affine transformations preserve both parallel lines and ratios between distances of collinear points. It thus suffices to show the desired result when $ABCD$ is a square. This can be done in a variety of ways. For instance, a coordinate bash can be applied by setting $A$ to be the origin. Let the length of the square be $1$ and set $X$ and $Y$ as $(a, 0)$ and $(0, b)$ respectively, so the line $XY$ has equation $bx + ay = ab$. Then, we note that $Z$ is the point $\left(\frac{ab}{a+b}, \frac{ab}{a+b}\right)$, so
$$\frac{AB}{AX} + \frac{AD}{AY} = \frac{1}{a} + \frac{1}{b} = \frac{a+b}{ab} = \frac{AC}{AZ}$$