- (Method 1) As shown in Figure 3, take point $F$ on side $AC$ such that $AF = AB$, then $FC = \frac{\sqrt{2}}{2} BC$, thus $FC^2 = \frac{1}{2} BC^2$.

Take the midpoint $M$ of $BC$, then $\frac{1}{2} BC^2 = BC \cdot MC$. That is, $FC^2 = BC \cdot MC$. Therefore, it is easy to know that $\triangle FMC \sim \triangle BFC$. Thus, $\angle AFB = \angle FMB$, and $\frac{BF}{FM} = \frac{BC}{FC} = \sqrt{2}$.

Extend $MF$ to intersect line $ED$ at point $N$. It is easy to know that $\triangle BMN$ is an isosceles triangle. Combining $\angle AFB = \angle FMB$, it is easy to know that $\triangle BMN \sim \triangle BFA$, i.e., $\triangle BAN \sim \triangle BFM$, hence $\frac{AB}{AN} = \frac{FB}{FM}$.

From $\angle BAF = \angle BNM$, we know that $B, A, N, F$ are concyclic, thus $\angle ANB = \angle AFB = \angle NMB = \angle NBM$. Therefore, $\angle AND = \angle ANB + \angle BNE = \angle NBM + \angle BNE = 90^\circ$. Therefore, in the right triangle $\triangle AND$, $\frac{AD}{AN} = \frac{AB}{AN} = \frac{FB}{FM} = \sqrt{2}$, $\triangle AND$ is an isosceles right triangle, thus $\angle AND = 45^\circ \Rightarrow \angle ADE = 135^\circ$.
$$
\frac{3ab}{3a + 2\sqrt{2}c}.
$$

In fact, we have
$$
\begin{aligned}
CK & = \frac{3a}{4} \cdot \frac{2ab}{a^2 + b^2 - c^2} = \frac{3a^2b}{2(a^2 + b^2 - c^2)} = 3ab \cdot \frac{\sqrt{2}(b - c)}{2(2b^2 + 2c^2 - 4bc + b^2 - c^2)} \\
& = 3ab \cdot \frac{\sqrt{2}(b - c)}{2(3b^2 + c^2 - 4bc)} = 3ab \cdot \frac{\sqrt{2}(b - c)}{(3\sqrt{2}b - \sqrt{2}c) \cdot \sqrt{2}(b - c)} = \frac{3ab}{3\sqrt{2}b - \sqrt{2}c} \\
& = \frac{3ab}{3a + 2\sqrt{2}c}.
\end{aligned}
$$

Therefore, $\angle EDF = 45^\circ \Rightarrow \angle ADE = 135^\circ$.
As shown in Figure 1, in $\triangle A B C$, $A C-A B=\frac{\sqrt{2}}{2} B C$. Point $D$ is a point inside $\triangle A B C$ such that $A D=A B$. A perpendicular line is drawn from point $D$ to $B C$, with the foot of the perpendicular being $E$. If $3 B E=C E$, find the measure of $\angle A D E$.