When two parallel lines are cut by a transversal, alternate angles formed are equal.
In optics the angle of incidence (the angle which an incident ray makes with a perpendicular to the surface at the point of incidence) and the angle of reflection (the angle formed by the reflected ray with a perpendicular to the surface at the point of incidence) are equal.
Draw a perpendicular line BL and CM at the point of incidence on both mirrors. Since PQ and RS are parallel to each other, perpendiculars drawn are also parallel i.e, BL || CM.
Since BC is a transversal to line BL and CM, alternate interior angles are equal.
∴ ∠LBC = ∠BCM = x .......(1)
By first law of reflection :
The angle of incidence (the angle which an incident ray makes with a perpendicular to the surface at the point of incidence) and the angle of reflection (the angle formed by the reflected ray with a perpendicular to the surface at the point of incidence) are equal.
By law of reflection, at the first point of incidence B on mirror PQ, we get :
⇒ ∠ABL = ∠LBC = x
⇒ ∠ABC = ∠ABL + ∠LBC
⇒ ∠ABC = x + x
⇒ ∠ABC = 2x ......(2)
By law of reflection, at the second point of incidence C on mirror RS, we get :
⇒ ∠MCD = ∠BCM = x
From figure,
⇒ ∠BCD = ∠BCM + ∠MCD
⇒ ∠BCD = x + x
⇒ ∠BCD = 2x .......(3)
From equation (2) and (3), we get ∠ABC = ∠BCD which are alternate interior angles for the lines AB and CD and BC as the transversal.
We know that, if a transversal intersects two lines such that a pair of alternate interior angles are equal, then the two lines are parallel.
Since, alternate interior angles are equal, we can say AB is parallel to CD (AB || CD).
Hence, proved that AB || CD.