As shown in the figure, in $\triangle A B C$, $\angle A B C=90^{\circ}, \quad B C=8$. $D$ and $E$ are points on $A B$ and $A C$ respectively, such that $C D=D E$, $\angle D C B=\angle E D A$. If the area of $\triangle E D C$ is 50, then the length of $A B$ is $\qquad$.
Geometry; Pythagorean Theorem, Chord Diagram; Similar Triangles, Pyramid Model.
Draw $E F \perp A B$, with the foot of the perpendicular at $F$.
In $\triangle B C D$ and $\triangle F D E$,
$$
\left\{\begin{array}{c}
\angle C B D=\angle D F E=90^{\circ} \\
\angle D C B=\angle E D F \\
C D=D E
\end{array}\right.
$$

Therefore, $\triangle B C D \cong \triangle F D E$;
(half of the chord diagram)
Thus, $B C=F D, B D=F E, \angle B C D=\angle F D E$;
Since $\angle B=90^{\circ}$;
Therefore, $\angle B C D+\angle B D C=90^{\circ}$;
Thus, $\angle F D E+\angle B D C=90^{\circ}$;
Since $\angle F D E+\angle C D E+\angle B D C=180^{\circ}$;
Therefore, $\angle C D E=90^{\circ}$;
This way, we can conclude that $\triangle E D C$ is an isosceles right triangle;
And from $S_{\triangle E D C}=50$, we get $C D=D E=10$;
And $B C=F D=8$;
By the Pythagorean Theorem, we get $B D=F E=6$.

In $\triangle A B C$ and $\triangle A F E$,
$$
\left\{\begin{array}{c}
\angle B A C=\angle F A E \\
\angle A B C=\angle A F E=90^{\circ}
\end{array}\right. \text {; }
$$

Therefore, $\triangle A B C \sim \triangle A F E$;
(pyramid model);
Thus, $\frac{E F}{B C}=\frac{A F}{A B}$;
$$
\frac{6}{8}=\frac{A B-14}{A B} \text {; }
$$

Solving, we get $A B=56$.