Ray in red is the shortest path if and only if $\theta_I=\theta_R$ and light only travels in such path to reach point B via mirror.
It is not a local minimum, because the shortest way, of course is a straight line connecting $A$ and $B$.
Red line with $\theta_I = \theta_R$ is a stationary curve of the group of curves that go through A, B and a point (s) on the mirror.
In an ellipsoidal case, you have three points
What is the shortest path connects $O$,$D$ and back to $O$? of course, it is a straight line going from $O$ to $D$ and then back to $O$. Although the time travels from $O$ to $O$ is maximum, for the particular path, $O$-$D$-$O$, light chooses the path such that time it travels is the minimum.
In the mirror example, the problem that is in question is not the minimum path between two points ($A$ and $B$). The real question is
Where is a point in the mirror $P$, such that if you connext $A$-$P$-$B$, it gives you the minimum distance?
Then you can make an argument as represented in wikibook page to find the point $P$.
Turns out, empirically, we have enough data and examples to conclude that light travels such path (the path such that time the light travels is the least).
Yes, in the mirror example, light source can be a light bulb, and the light from the light bulb can travel completely opposite to B, reflected off the wall of the room, bounced off 1000000 times, and finally reach the point B. Sure this way takes the most time.
However, the path that light have taken is the path such that connects all 1000000 points of reflection and gives light a least time to travel.
Hope it helps.