5. (10 points) In isosceles trapezoid $A B C D$, $A B$ is parallel to $C D, A B=6, C D=14, \angle A E C$ is a right angle, $C E$ $=C B$, then $A E^{2}$ equals ( )
A. 84
B. 80
C. 75
D. 64
【Analysis】As shown in the figure, connect $A C$, draw $A F \perp C D$ at point $F$, and draw $B G \perp C D$ at point $G$, constructing right triangles $\triangle A F C$ and $\triangle B G C$. Use the Pythagorean theorem to find the value of $A E^{2}$.

【Solution】Solution: As shown in the figure, connect $A C$, draw $A F \perp C D$ at point $F$, and draw $B G \perp C D$ at point $G$, then $A F=B G, A B=F G$ $=6, D F=C G=4$.
In the right triangle $\triangle A F C$, $A C^{2}=A F^{2}+F C^{2}=A F^{2}+10^{2}=A F^{2}+100$,
In the right triangle $\triangle B G C$, $B C^{2}=B G^{2}+G C^{2}=A F^{2}+4^{2}=A F^{2}+16$,
Also, since $C E=C B, \angle A E C=90^{\circ}$,
Therefore, $A E^{2}=A C^{2}-E C^{2}=A F^{2}+100-\left(A F^{2}+16\right)=84$, that is, $A E^{2}=84$.
Hence, the answer is: A.