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 $DM^{2} = DN \cdot MC.$

(N/A) Join $DB.$
We have $DN \parallel CB,$ $DM \parallel AB,$ and $\angle B = 90^{\circ}.$
Therefore,$DMBN$ is a rectangle.
Thus,$DN = MB$ and $DM = NB.$
Since $D$ is the foot of the perpendicular drawn from $B$ to $AC,$ we have $\angle CDB = 90^{\circ}.$
$\angle 2 + \angle 3 = 90^{\circ} \dots(1)$
In $\Delta CDM, \angle 1 + \angle 2 + \angle DMC = 180^{\circ} \Rightarrow \angle 1 + \angle 2 = 90^{\circ} \dots(2)$
In $\Delta DMB, \angle 3 + \angle DMB + \angle 4 = 180^{\circ} \Rightarrow \angle 3 + \angle 4 = 90^{\circ} \dots(3)$
From $(1)$ and $(2),$ we get $\angle 1 = \angle 3.$
From $(1)$ and $(3),$ we get $\angle 2 = \angle 4.$
In $\Delta DCM$ and $\Delta BDM,$
$\angle 1 = \angle 3$ (Proved above)
$\angle 2 = \angle 4$ (Proved above)
Therefore,$\Delta DCM \sim \Delta BDM$ ($AA$ similarity criterion).
Thus,$\frac{BM}{DM} = \frac{DM}{MC}.$
Since $BM = DN,$ we have $\frac{DN}{DM} = \frac{DM}{MC}.$
Therefore,$DM^{2} = DN \cdot MC.$