[tex3]\mathsf{Trace ~ RH\perp AB\\
∠MAP=∠BAC \\ AB=AM, M,N,E (collneares).\\
△ARH∼△PAB ~e ~ RH=AB.\\
[ARB]=[AMN](1)\\
\therefore △APB∼△EPN⟹PBAB=PNEN\\
PB⋅EN=AB⋅PN⟹[EPB]=[MPN](2)\\
(1)+(2),\\
[ARB]+[EPB]=S1\\
[ARB]+[EPB]+[ERP]=S1+S2\\
Mas [ARB]+[EPB]+[ERP]=[ARE]−[ABP] =[ACP]−[ABP]=[ABC]}\\
\therefore \boxed{\color{red}{ABC} = 9+6 =15m^2}[/tex3]
(Solução:MathLover)
Outra solução trigonométrica -
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