In $\Delta XYZ$,$m\angle Y = 90^{\circ}$ and $\overline{YM}$ is an altitude to the hypotenuse $\overline{XZ}$. If $XM = 16ZM$,prove that $XY = 4YZ$.

(N/A) In $\Delta XYZ$,$\angle Y = 90^{\circ}$ and $\overline{YM} \perp \overline{XZ}$.
According to the Geometric Mean Theorem (or properties of similar triangles),in a right-angled triangle,the altitude to the hypotenuse creates two triangles similar to the original triangle and to each other.
Specifically,$\Delta XMY \sim \Delta YMZ$.
From the similarity $\Delta XMY \sim \Delta YMZ$,we have the ratio of corresponding sides:
$\frac{XY}{YZ} = \frac{XM}{YM} = \frac{YM}{ZM}$.
From $\frac{XY}{YZ} = \frac{YM}{ZM}$,we get $YM^2 = XY \cdot ZM$ (not directly useful here).
Alternatively,using the property of the altitude in a right triangle:
$XY^2 = XM \cdot XZ$ and $YZ^2 = ZM \cdot XZ$.
Dividing these two equations:
$\frac{XY^2}{YZ^2} = \frac{XM \cdot XZ}{ZM \cdot XZ} = \frac{XM}{ZM}$.
Given $XM = 16ZM$,we substitute this into the ratio:
$\frac{XY^2}{YZ^2} = \frac{16ZM}{ZM} = 16$.
Taking the square root of both sides:
$\frac{XY}{YZ} = \sqrt{16} = 4$.
Therefore,$XY = 4YZ$.