2. As shown in Figure 2, a circle with radius 1 passes through vertex $A$ of $\triangle A B C$, and is tangent to side $B C$ at point $D$, intersecting sides $A B$ and $A C$ at points $E$ and $F$, respectively. If $E F$ bisects $\angle A F D$, and $\angle A D C=80^{\circ}$, does there exist a triangle that satisfies the conditions such that $\frac{A B+B C+C A}{A D^{2}}$ is an irrational number, and this irrational number is a root of a quadratic equation with integer coefficients? If not, please prove it; if so, please find a point that satisfies the conditions and calculate the value.
2. Existence.

It is known that $\angle A E D=\angle A D C=80^{\circ}$,
$$
\begin{array}{c}
\angle A F D=100^{\circ}, \\
\angle B A D=\angle E F D=50^{\circ}, \\
\angle A B D=30^{\circ},
\end{array}
$$

and $A D=2 \sin 80^{\circ}$.
Let $\angle A C D=\alpha$. Then $20^{\circ}<\alpha<80^{\circ}$.
$$
\begin{array}{l}
\text { By } \frac{A B}{A D}=\frac{\sin 100^{\circ}}{\sin 30^{\circ}}=2 \sin 100^{\circ}, \\
\frac{A C}{A D}=\frac{\sin 80^{\circ}}{\sin \alpha}, \\
\frac{B C}{A D}=\frac{B D}{A D}+\frac{D C}{A D}=2 \sin 50^{\circ}+\frac{\sin \left(80^{\circ}+\alpha\right)}{\sin \alpha}, \\
\frac{A B+B C+C A}{A D} \\
=2 \sin 100^{\circ}+\frac{\sin 80^{\circ}}{\sin \alpha}+2 \sin 50^{\circ}+ \\
\quad \frac{\sin 80^{\circ} \cdot \cos \alpha}{\sin \alpha}+\cos 80^{\circ} \\
=2 \sin 100^{\circ}+2 \sin 50^{\circ}+\cos 80^{\circ}+ \\
\sin 80^{\circ} \cdot \frac{1+\cos \alpha}{\sin \alpha} .
\end{array}
$$

We know that $\frac{A B+B C+C A}{A D}$
Let $\alpha=60^{\circ}$. Then
$$
\frac{A B+B C+C A}{A D}
$$
$$
\begin{array}{l}
=2 \sin 100^{\circ}+2 \sin 50^{\circ}+\cos 80^{\circ}+\sqrt{3} \sin 80^{\circ} \\
=2 \sin 100^{\circ}+2 \sin 50^{\circ}+2 \sin 110^{\circ} \\
=2 \sin 80^{\circ}+4 \sin 80^{\circ} \cdot \cos 30^{\circ} \\
=2(1+\sqrt{3}) \sin 80^{\circ} . \\
\text { Hence } \frac{A B+B C+C A}{A D^{2}}=1+\sqrt{3} \text { (irrational number). }
\end{array}
$$