= Solution
A plane <tree> has one <face of a plane graph>, so its <dual graph> has one <vertex>. Every original <edge> is a bridge incident to that same face on both sides, so every dual <edge> is a loop. Thus the dual drawing in the figure is a bouquet of eight loops, arranged with the displayed nesting inherited from the <tree>. It has nine <faces of a plane graph>, matching $f^*=n=9$. The nesting matters for which faces meet; a bouquet drawn as eight unrelated petals would have a different face-adjacency <graph>.
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