Example 7 Given that the three sides $a, b, c$ and the three interior angles $A, B, C$ of triangle $ABC$ satisfy the condition: $\frac{c}{\cos \frac{C}{2}}=a \cdot \operatorname{ctg} B+b \cdot \operatorname{ctg} A$, determine the shape of this triangle.
Solution: From the given conditions and the Law of Sines, we have
$2 \sin \frac{c}{2}=\frac{\sin A}{\sin B} \cos B+\frac{\sin B}{\sin A} \cos A$,
which means $\sqrt{2(1-\cos C)}=\frac{a}{b} \cos B+\frac{b}{a} \cos A$.

By the Law of Cosines, we get
$$
\sqrt{2-\frac{a^{2}+b^{2}-c^{2}}{a b}}=\frac{a}{b} \cdot \frac{a^{2}+c^{2}-b^{2}}{2 a c}+\frac{b}{a} \cdot \frac{b^{2}+c^{2}-a^{2}}{2 b c} .
$$

Squaring both sides of the above equation and simplifying, we get
$$
(a-b)^{2}\left[(a-b)^{2}(a+b)^{2}+2 c^{2}\left(a^{2}+b^{2}+4 a b\right)+c^{4}\right]=0 .
$$

Since $c^{4}+2\left(a^{2}+b^{2}+4 b c\right) c^{2}+(a+b)^{2}(a-b)^{2}>0$, simplifying yields $(a-$ $b)^{2}=0$, i.e., $a=b$. Therefore, $\triangle A B C$ is an isosceles triangle.