# mirror reflection

1. The problem statement, all variables and given/known data

Question A. Draw accurately in the figure the light beam that goes from Object A to eye B after being reflected on the mirror. It must be consistent with the mirror principle!

Question B. At question A. you may have connected the point behind the mirror (A') with eye B. And found the correct light path that way.

Will that also work if you use the point behind the mirror of B (B') instead? Do you get the same result? Use mathematical arguments for your judgement!

2. Relevant equations mirror reflection rules

3. The attempt at a solution

Question A: I tried to draw point A' and connected that to B. The purple line is the normal line (the imaginary line that is perpendicular to the mirror) And the angle between the mirror and A is the same as the angle between the mirror and B.

Question B: I think this is true, I can draw the same lines, but I don't know what is meant by 'mathematical arguments'

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This is a very nicely written question. If only we had more like this one... :-) – Vedran Šego Nov 14 '13 at 4:14

Notice that $AA'B'B$ is an isosceles trapezoid, so its diagonals intersect on the mirror, which means that using either $\triangle AA'B$ or $\triangle AB'B$ will give you the same point.

You'll probably need to provide a bit more arguments that the diagonals intersect on the mirror. Read the linked article and ask if you get stuck.

Edit: Here is a bit more:

We can extend $AA'B'B$ to a triangle. Let us denote the new vertex as $C$. It is on the intersection of the lines on which $\overline{AB}$ and $\overline{A'B'}$ lie.

Note that

1. $|\overline{AA''}| = |\overline{A'A''}|$, where $A''$ denotes a point at which $\overline{AA'}$ intersects with the mirror; and
2. the mirror (on which the altitude of the mirror lies) is orthogonal to $\overline{AA'}$.

From points $1$ and $2$, we see that $\triangle AA'C$ is an isosceles triangle. The same can be said for $\triangle BB'C$ (with the same arguments).

Now, since

$$|\overline{AB}| = |\overline{AC}| - |\overline{BC}| = |\overline{A'C}| - |\overline{B'C}| = |\overline{A'B'}|,$$

we see that $AA'B'B$ is an isosceles trapezoid.

Using this, show that the diagonals of $AA'B'B$ intersect on the mirror. Can you take over now?

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Do you mean with a bit more arguments to proof Similarity (geometry) between the triangles? – user108681 Nov 14 '13 at 4:36
@user108681 I've expanded my answer. – Vedran Šego Nov 14 '13 at 12:15
A bit easier: you can just observe isosceles triangles $\triangle AA'E$ and $\triangle BB'E$, where $E$ is the point on the mirror where $\overline{AB'}$ and $\overline{A'B}$ intersect (i.e., the point you're looking for). – Vedran Šego Nov 14 '13 at 18:31