Let $\triangle ABC$ be an isosceles triangle with $AB = AC$ and $BC = 14$. This triangle is inscribed in a circle $\omega$. Let $D$ be a point on the ray $BC$ such that $CD = 6$. The intersection of line $AD$ with the circle $\omega$ is point $E$. Given that $AE = 7$, find $AC^2$.

Let's think step by step and output the final answer within \boxed{}.
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