Geometric Mirror Problem with light beam

Question

Two mirrors (represented by the blue lines) share an acute angle in the Point $A$. A beam of light $Y$ that is parallel to the horizontal mirror is reflected in the Point $C$. Then it is again reflected in $E$ and in $F$ and touches the horizontal mirror in a right angle in $G$.

We know that $AG$ has the length $5$, so which of the options is the lenght of $m=GH$?

1. $4 \sqrt{2}$
2. $5\sqrt{2}$
3. $10$

My Progress:

We label the angles and because of the law of regular reflection and the parallelity of $Y$ to the horizontal mirror we get:

Now my thinking was, that $m$ could be any value that is bigger $FG$, because we can make $\alpha$ as big as we want, as long as it is acute. Where am I wrong and how do I find the right solution?

Answer 1

If $\small \angle GAF = \alpha$, $\small \angle HCE = 90^0 - 2 \alpha \implies \angle GEF = \angle HEC = 2\alpha$

and $\small \angle GFE = 90^0 - 2 \alpha \implies \angle AFG = \angle CFE = 45^0 + \alpha$

So we have, $\small \alpha = \displaystyle \frac{\pi}{8}$

Now $\small GF = AG \tan \alpha, GE = GF \cot 2\alpha, FE = GF \csc 2\alpha$

As $\small \angle CEF = 90^0, CE = FE \tan (45^0+\alpha), EH = CE \cos 2\alpha$

Simplifying, $\small GH = GE + EH = AG (1 + \tan \alpha) = 5 \sqrt2$

Answer 2

Applying exterior angle theorem gives $\delta=\alpha+\alpha=2\alpha$ and similarly proceeding will give $\beta=3\alpha$. $\alpha+\beta=90^{\circ}$ and thereafter $\alpha=\left(\frac{45}{2}\right)^{\circ}$. $FG$ can be found from $\triangle AGF$.

$\triangle FGE$ and $\triangle CHE$ are both right isosceles triangles. Hence, $AH=AG+GE+EH=AG+FG+GH$.

$\triangle AGF\sim \triangle AHC$ and thus $\frac{AG}{AH}=\frac{FG}{CH}$. Put the values of $AG$ and $FG$ alongside $m=CH$ and solve for it .