Doesn't the eutectic diagram violate Gibbs's phase rule?

EDIT: As the question hasn't attracted any answers, I tried to put in more detail.


Gibbs's phase rule for a 2 component system at constant pressure says that $F=3-P$, i.e., that the number of degrees of freedom is three minus the number of phases present. This should mean that when $P=3$, there should be no degrees of freedom and three phases should only be in equilibrium at a single point.


However, looking at the eutectic phase diagram, this seems not to be true. Even visually - four lines intersect at the eutectic point, not three.


[![enter image description here](https://i.stack.imgur.com/u02q7.png)](https://i.stack.imgur.com/u02q7.png)


And my understanding is that the three phases, $\alpha$, $\beta$ and $L$ can be in equilibrium along a line at the eutectic temperature, not only at a single point.


This can be seen by looking at the Gibbs free energy of the three phases at the eutectic temperature, which should look like this:


[![enter image description here](https://i.stack.imgur.com/m2CiH.png)](https://i.stack.imgur.com/m2CiH.png)


By the common tangent construction, the three phases are at equilibrium along a range of compositions, and thus $F=1$, not $0$.


Is there a mistake in my understanding, or is it a 'violation' of the phase rule?




The three-phase (eutectic) equilibrium phase compositions are not along a line in the phase diagram. Each of the three phases is instead at a single point (specifically, where the isotherm touches the field having that phase), which dovetails with zero degrees of freedom. The full isothermal line is not a range of equilibrium compositions but a boundary between different two-phase regions above and below that temperature. In contrast, other lines actually do represent a range of equilibrium compositions when the necessary degree of freedom (shown in the diagram as the temperature being allowed to vary as a function of composition) exists.


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