Simple geometric proof for Snell's law of refraction

Snell's law of refraction can be derived from Fermat's principle that light travels paths that minimize the time using simple calculus. Since Snell's law only involves sines I wonder whether this minimum problem has a simple geometric solution.

Perhaps this will help, if you are looking at a non Calculus approach. Consider two parallel rays $A$ and $B$ coming through the medium $1$ (say air) to the medium $2$ (say water). Upon arrival at the interface $\mathcal{L}$ between the two media (air and water), they continue their parallel course in the directions $U$ and $V$ respectively.

Let us assume that at time $t=0$, light ray $A$ arrives at the interface $\mathcal{L}$ at point $C$, while ray $B$ is still shy of the surface by a distance $PD$. $B$ travels at the speed $v_{1}=\frac{c}{n_{1}}$ and arrives at $D$ in $t$ seconds. During this time interval, ray $A$ continues its journey through the medium $2$ at a speed $v_{2}=\frac{c}{n_{2}}$ and reaches the point $Q$.

We can formulate the rest, geometrically (looking at the parallel lines) from the figure. Let $x$ denote the distance between $C$ and $D$. \begin{eqnarray*} x \sin\left(\theta_{i}\right) &=& PD \\ &=& v_{1} t \\ &=& \frac{c}{n_{1}} t \\ x \sin\left(\theta_{r}\right) &=& CQ \\ &=& v_{2} t \\ &=& \frac{c}{n_{2}} t \end{eqnarray*}

Thus,

\begin{eqnarray*} n_{1} \sin\left(\theta_{i}\right) &=& \frac{c}{x} t \\ n_{2} \sin\left(\theta_{r}\right) &=& \frac{c}{x} t \end{eqnarray*}

Re arranging this will take us to the Snell's law as we know. \begin{eqnarray*} \frac{n_{2} }{n_{1}} &=& \frac{\sin\left(\theta_{i}\right) }{ \sin\left(\theta_{r}\right)} \end{eqnarray*}

Let upper half plane be a medium with refractive index $n_1$and the lower half plane be another medium with refractive index $n_2$. Let $(a_1,b_1)$and $(a_2,b_2)$ be the starting point and end point of light traveled, while $(x,0)$ is the point of refraction.

Time of light traveled $$t=\frac {n_1}c\sqrt{(x-a_1)^2+b_1^2}+\frac {n_2}c\sqrt{(a_2-x)^2+b_2^2}$$ where $c$ is the speed of light

By Fermat's principle, we need to minimize $t$ $$\frac{dt}{dx}=0$$$$\frac {n_1}c \frac {x-a_1}{\sqrt{(x-a_1)^2+b_1^2}}+\frac {n_2}c \frac {-x+a_2}{\sqrt{(a_2-x)^2+b_2^2}}=0$$$$n_1\frac {x-a_1}{\sqrt{(x-a_1)^2+b_1^2}}=n_2\frac {x-a_2}{\sqrt{(a_2-x)^2+b_2^2}}$$ note that $\frac {x-a_1}{\sqrt{(x-a_1)^2+b_1^2}}=\sin\theta_1$ and $\frac {x-a_2}{\sqrt{(a_2-x)^2+b_2^2}}=\sin\theta_2$, proved.