# ListLinePlot not showing one label in a cycle of points

I am using ListLinePlot to represent a closed circuit of points (forming a polyhedron). I am trying to label each node (point) of the cycle, but the node that is present twice gets no label.

This is my example:

ListLinePlot[{{11, 7}, {7, 9}, {15, 15}, {16, 12}, {11, 7}} -> {"A'",
"B'", "C'", "D'", " "},
LabelingFunction ->
Callout[Automatic, Automatic, Appearance -> None],
Filling -> Bottom, AspectRatio -> Automatic,
PlotRange -> {{0, 20}, {0, 20}}, AxesLabel -> {"z1", "z2"},
Mesh -> Full,
GridLines -> Function[{min, max}, Range[Floor[min], Ceiling[max]]]]


This is the result I get:

I think the problem is with point {11, 7} (that should be labelled with "A'"), which appears twice in the main argument of the function ListLinePlot. I tried different approaches, but nothing seemed to work.

Any assistance will be appreciated.

Since you don't seem to want the stem / leader feature from Callout, try LabelingFunction -> Labeled instead. The results are practically identical to what you show in your answer:

ListLinePlot[
{{11, 7}, {7, 9}, {15, 15}, {16, 12}, {11, 7}} -> {"A'", "B'", "C'", "D'", ""},

LabelingFunction -> Labeled,

Filling -> Bottom,
AspectRatio -> Automatic,
PlotRange -> {{0, 20}, {0, 20}},
AxesLabel -> {"z1", "z2"},
Mesh -> Full,
GridLines -> Function[{min, max}, Range[Floor[min], Ceiling[max]]]
]


I found a way, using Show, which allows overlaying different plots:

Show[
ListLinePlot[{{11, 7}, {7, 9}, {15, 15}, {16, 12}, {11, 7}} -> {"A'",
"B'", "C'", "D'", "-"},
LabelingFunction ->
Callout[Automatic, Automatic, Appearance -> None],
Filling -> Bottom, AspectRatio -> Automatic,
PlotRange -> {{0, 20}, {0, 20}}, AxesLabel -> {"z1", "z2"},
Mesh -> Full,
GridLines -> Function[{min, max}, Range[Floor[min], Ceiling[max]]]],
ListLinePlot[{{11, 7}} -> {"A'"},
LabelingFunction ->
Callout[Automatic, Automatic, Appearance -> None],
PlotRange -> {{0, 20}, {0, 20}}]
]


I get this:

Anyway, I think a solution without using the Show hack should be possible.