Why are points on the axis of a circuit with axial symmetry equipotential? Given the following electrical circuit:

How to explain that L and K are equipotential?
More generic question: why given any electrical circuit with axial symmetry, points that belong to the axis are equipotential:

P.S. It's derived from my more complex homework, but I am trying to understand how things work, not just trying to get it solved.
Update. There is a resistance between K and L instead of ideal wire.
 A: Assuming the wire is ideal and since there are no resistances between the point L and K, the potential drop between L and K is zero i.e potential does not change. So the potential is same anywhere in between those two points and therefore, they are equipotential.
UPDATE:


Mathematical Proof: Assume potential at A, L, K and B are A, x, y and B. Using Kirchhoff's law we get:
$$\frac{y-B}{R}+\frac{y-x}{4R}+\frac{y-0}{R}=0$$
$$\frac{x-0}{2R}+\frac{x-y}{4R}+\frac{x-B}{2R}=0$$
Upon addition of the above equations and further simplification, we get $x=\frac{B}{2}$ and $y=\frac{B}{2}$. This means the potential at L and K are same i.e equipotential and the existence of the 4R resistor does not matter.
However, it is very difficult to use Kirchhoff's law every time, especially when you have something as ambiguous as a circuit like this:

For these type of circuits, we must find symmetricity. However, note that if there is no symmetricity then Kirchhoff's law is the only way to solve this. Luckily there is symmetricity here and I can guide you through the thought process. Let us take the first diagram as an example:

Let us say $i$ current flows in through A which breaks as $i_1$ in wire AK and $i_2$ in wire AL. Now since the wires connected to B have similar resistances as the wires connected to A, the same amount of current must flow in wire LB as in AL and exit through B and the same amount of current must flow in the wire KB as in AK and exit through B. So now we can notice that the $i_1$ current in AK does not enter the wire LK and the $i_2$ current in AL does not enter the wire LK. This means, there is no flow of charge or current through the wire LK. 
