Chapter 14.2, Problem 6E

To determine

To prove: that the light ray follows the shortest possible path from A to C via the mirror by proving that for any point E on the mirror (other than B ) $AE+EC>AB+BC$ .

Explanation of Solution

Given information:

The given statement is that the condition can be proved by proving that for any point E on the mirror (other than B ) $AE+EC>AB+BC$ .

Concept used:

The triangle inequality

The sum of lengths of any two sides of a triangle is greater than length of third side.

Proof:

The geometric diagram to proof the condition.

Since triangle BDO is reflection of triangle ABOin the line EOand $AO=OD$ therefore $AB=BD$ .

And triangle EDO is reflection of triangle AEO in the line EO and $AO=OD$ therefore $AE=ED$

In triangle CDE

$\begin{array}{l}EC+ED>CD\left[\text{triangle inequality}\right]\\ EC+ED>BC+BD\left[CD=BC+BD\right]\\ EC+AE>BC+AB\left[AE=ED,BD=AB\right]\\ AE+EC>AB+BC\end{array}$

The given condition is proved.

Therefore,

It is proved that light ray follows shortest path.