Question

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

Answer 1

Given: Two plane mirrors m and n, perpendicular to each other. CA is incident ray and BD is reflected ray.
To Prove: $CA\parallel DB$
Construction: OA and OB are perpendiculars to m and n respectively.

Proof:

$\because m\perp n,OA\perp m$ and $OB\perp n$

$\therefore \angle AOB={90}^o$

(Lines perpendicular to two perpendicular lines are also perpendicular.)

In $\Delta AOB,$

$\angle AOB+\angle OAB+\angle OBA={180}^o$

$\Rightarrow {90}^o+\angle 2+\angle 3={180}^o\Rightarrow \angle 2+\angle 3={90}^o$

$\Rightarrow 2(\angle 2+\angle 3)={180}^o$ (Multiplying both sides by 2)

$\Rightarrow 2\left(\angle 2\right)+2\left(\angle 3\right)={180}^o$

$\Rightarrow \angle CAB+\angle ABD={180}^o$

(Angle of incidence $=$ Angle of reflection)

$\therefore \angle 1=\angle 2$ and $\angle 3=\angle 4)$

$\Rightarrow CA\parallel BD$ ($\angle CAB$ & $\angle ABD$ form a pair of consecutive interior angles and are supplementary)