In the figure,$D$ is a point on the hypotenuse $AC$ of $\Delta ABC,$ such that $BD \perp AC,$ $DM \perp BC,$ and $DN \perp AB.$ Prove that $DN^{2} = DM \cdot AN.$

(N/A) In quadrilateral $DMBN,$ since $\angle DMB = 90^{\circ},$ $\angle DNB = 90^{\circ},$ and $\angle MBN = 90^{\circ}$ (given $\angle ABC = 90^{\circ}$),the fourth angle $\angle MDN$ must also be $90^{\circ}.$ Thus,$DMBN$ is a rectangle.
Therefore,$DM = NB$ and $DN = MB.$
In $\Delta ABC,$ $BD \perp AC.$ By the property of similar triangles in a right triangle,$\Delta ADN \sim \Delta ABD$ and $\Delta BDC \sim \Delta ABD.$
More directly,in $\Delta DNB$ and $\Delta ADN:$
$\angle DNB = \angle DNA = 90^{\circ}.$
$\angle ND B = \angle DAN$ (since both are complementary to $\angle ADN$).
Thus,$\Delta DNB \sim \Delta ADN$ by $AA$ similarity.
Therefore,$\frac{DN}{AN} = \frac{NB}{DN}.$
$DN^{2} = AN \cdot NB.$
Since $NB = DM,$ we have $DN^{2} = AN \cdot DM.$