\(\begin{align}
  & \cos C=\frac{{{a}^{2}}+{{b}^{2}}-{{c}^{2}}}{2ab}=\frac{{{a}^{2}}+{{b}^{2}}-\frac{{{a}^{2}}+{{b}^{2}}}{3}}{2ab}=\frac{1}{3}\times \frac{{{a}^{2}}+{{b}^{2}}}{ab}\ge \frac{2}{3} \\
& \sin C=\sqrt{1-{{\cos }^{2}}C}=\frac{\sqrt{5}}{3} \\
\end{align}\)

設AC和BD交於E
\(\begin{align}
  & \frac{AE}{CE}=\frac{\Delta ABD}{\Delta CBD}=\frac{7\times 7}{5\times 3}=\frac{49}{15} \\
& \frac{BE}{DE}=\frac{\Delta ABC}{\Delta ADC}=\frac{7\times 5}{7\times 3}=\frac{5}{3} \\
\end{align}\)
剩下的就簡單了