In the figure,$m$ and $n$ are two plane mirrors perpendicular to each other. Show that the incident ray $CA$ is parallel to the reflected ray $BD$.

(N/A) Let the normals at $A$ and $B$ meet at $P$.
Since the mirrors are perpendicular to each other,$BP \parallel OA$ and $AP \parallel OB$.
Thus,$BP \perp PA$,i.e.,$\angle BPA = 90^{\circ}$.
In $\triangle BPA$,by the angle sum property:
$\angle 2 + \angle 3 + \angle BPA = 180^{\circ}$
$\angle 2 + \angle 3 + 90^{\circ} = 180^{\circ}$
$\angle 2 + \angle 3 = 90^{\circ} ......(1)$
By the laws of reflection,the angle of incidence equals the angle of reflection:
$\angle 1 = \angle 2$ and $\angle 3 = \angle 4$
Substituting these into $(1)$:
$\angle 1 + \angle 4 = 90^{\circ} ......(2)$
Now,consider the sum of the angles $\angle CAB$ and $\angle DBA$:
$\angle CAB + \angle DBA = (\angle 1 + \angle 2) + (\angle 3 + \angle 4)$
$= 2\angle 2 + 2\angle 3 = 2(\angle 2 + \angle 3)$
$= 2(90^{\circ}) = 180^{\circ}$
Since the sum of the interior angles on the same side of the transversal $AB$ is $180^{\circ}$,the lines $CA$ and $BD$ must be parallel.
Therefore,$CA \parallel BD$.