Question 54, Given in $\triangle ABC$, the angles $A, B, C$ are opposite to the sides $a, b, c$ respectively. If $c-a=2a \cdot \cos B$, find the minimum possible value of $\frac{3a+c}{b}$.

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Question 54, Solution: As shown in the figure, draw $C E \perp A B$ at $E$, and take the symmetric point $D$ of $B$ with respect to $E$. Then $\overline{\mathrm{AD}}=\overline{\mathrm{AB}}+\overline{\mathrm{BD}}=c-2 a \cdot \cos B=a=C B=C D$, thus we know $\Varangle A B C=180^{\circ}-\Varangle A D C=2 \angle A$, and $b=2 a \cdot \cos A$. Therefore,
$$
\frac{3 a+c}{b}=\frac{c-a+4 a}{2 a \cdot \cos A}=\frac{2 a \cdot \cos B+4 a}{2 a \cdot \cos A}=\frac{\cos 2 A+2}{\cos A}=\frac{2 \cos ^{2} A+1}{\cos A} \geq 2 \sqrt{2}
$$

When $\cos A=\frac{\sqrt{2}}{2}$, i.e., $A=\frac{\pi}{4}$, the equality holds.
Note: The directed segments and directed angles are used because $\angle A B C$ can be either acute or obtuse.