[tex3]\angle LBC = \theta\\
a = MC, ~b = HL, ~n = HN ~e~ m = NC\\
ab = 54\\
\text{Preenchendo os ângulos} \rightarrow LB\parallel MN\\
Tales: \frac{x}{a}=\frac{m}{n}(I)\\
T.Menelao~ \triangle HBC - FC: \\
\cancel{BF}.m.b = \cancel{FC}.n.12
\frac{b}{12}=\frac{m}{n}(II)\\
\frac{x}{a}=\frac{b}{12} \implies \boxed{\color{red}x = 4,5}[/tex3]
(Solução: Fornecida por Geobson -
viewtopic.php?f=4&t=94214&hilit=MC.HL%3D54&start=20)