26. Answer: 10
Since $B D \| F G$ and $F G$ is tangent to the circle at $C$, we have
$$
\angle B C F=\angle C B E=\angle D C G=\angle B D C=\angle B A C .
$$

Furthermore
$$
\angle B E C=\angle B A C+\angle A B E=\angle C B E+\angle A B E=\angle A B C=\angle A C B .
$$

We can then conclude that $B E=B C=D C=4$. Also, $\triangle A B E$ is similar to $\triangle D C E$. If we let $A E=x$,
$$
\frac{D E}{D C}=\frac{A E}{A B} \Longrightarrow D E=\frac{2}{3} x .
$$

By the Intersecting Chord Theorem, $A E \cdot E C=B E \cdot E D$, i.e. $x(6-x)=4\left(\frac{2}{3} x\right)$, which gives $x=\frac{10}{3}$, so $3 A E=3 x=10$.
26. In the diagram below, $A B C D$ is a cyclic quadrilateral with $A B=A C$. The line $F G$ is tangent to the circle at the point $C$, and is parallel to $B D$. If $A B=6$ and $B C=4$, find the value of $3 A E$.