Por propriedade, BF será bissetriz do triângulo BNC
e FN = FC
[tex3]\mathsf{BF ~é~bissetriz~\angle AFC\\
\therefore AF/FC=25/7 (T.Bissetriz)\implies \triangle AFC ~é~notável(7k:24k:25k)\\
\therefore \angle FAC=16°\\
BF~ é~ bissetriz \therefore \angle NFB=\angle BFC=37 \\
\therefore NF=FC~ e~ NB= BC\\
NB \perp AF \implies NB=7 ~e~ AN=24\\
\angle MNA=53^o\\
\angle AMN=90^o+16^o\\
T. Ângulo ~Inscrito \implies \angle MAC=37^o \therefore AM=MC\\
Traçando~altura ~MH, H ~será~ ponto ~médio~ de~ AC \therefore HC = 16 \\
\angle HMC=53\implies \triangle HMC-notável(3k, 4k, 5k) \therefore cos37=\frac{16}{MC}\implies MC=20=AM\\
\triangle AMH: MH^2 = 20^2 - 16^2 \therefore MH =12 \implies S_{AMC} =\frac{12.32}{2}=192\\
\triangle ANB: HE.25 = 7.24 \therefore HE = \frac{168}{25} \implies S_{ANC} = \frac{32.\frac{168}{25}}{2}=\frac{16.168}{25}\\
(
S=S_{AMC}-S_{ANC}=\boxed{\color{red}\frac{2112}{25}}}[/tex3]
Soluçao:jvmago adaptada -
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