# Ex.6.2 Q6 Lines and Angles Solution - NCERT Maths Class 9

## Question

In the given figure, \(PQ\) and \(RS\) are two mirrors placed parallel to each other. An incident ray \(AB\) strikes the mirror \(PQ\) at \(B\), the reflected ray moves along the path \(BC\) and strikes the mirror \(RS\) at \(C\) and again reflects back along \(CD\).

Prove that \(AB \|CD\).

## Text Solution

**What is known?**

\(PQ \| RS\)

**What is unknown?**

To prove: \(AB \| CD\)

**Reasoning:**

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.

**Steps:**

Draw perpendicular lines \(BL\) and \(CM\) at the point of incident on both mirrors since \(PQ\) and \(RS\) parallel to each other, perpendiculars drawn are parallel \(BL \| CM\). Since \(BC\) is a transversal to lines \(BL\) and \(CM\), alternate angles are equal so we get

\[\angle LBC = \angle BCM = x \;(\text{say})\; \ldots ( 1 )\]

By laws of reflection, at the first point of incidence \(B\), we get:

\[\begin{align} \angle ABL & = \angle LBC = x \\ \therefore \angle ABC& = \angle ABL+ \angle LBC \\ & = x + x \\ \therefore \angle ABC & = 2 x \ldots \ldots \ldots ( 2 ) \end{align}\]

By laws of reflection, at the first point of incidence \(C\), we get:

\[\begin{align} \angle MCD & = \angle BCM = x \\ \therefore \angle BCD& = \angle BCM + \angle MCD \\ & = x + x \\ \angle BCD &= 2 x \ldots \ldots ( 3 ) \end{align}\]

From equations (\(2\)) and (\(3\)), we get \(\angle ABC = \angle BCD\).

We know that, if a transversal intersects two lines such that a pair of alternate interior angles is equal, then the two lines are parallel. As Alternate angles are equal we can say \(AB \| CD\).