Let $AC=a$ and $AB=r$. By Pythagorean Theorem, $a^2+r^2=36^2$. Extending $\overline{CA}$ to intersect the circle again at $X$, $CX=a+r$. So by Power of a Point, $CE*CX=CD*CB\Rightarrow (a+r)(a-r)=a^2-r^2=16*36$. $a^2=\frac{(a^2+r^2)+(a^2-r^2)}{2}=\frac{36(16+36)}{2}=36*26=31^2-5^2=\boxed{936}$.

Proof 1. I"ll use the Stewart's relation: $20\cdot b^2+16\cdot c^2=36\cdot c^2+20\cdot 16\cdot 36\iff$ $20\cdot b^2=20\cdot c^2+20\cdot 16\cdot 36\iff$ $b^2=c^2+16\cdot 36\implies$ $\left\{\begin{array}{cc} b^2+c^2=36\cdot 36\\\\ b^2-c^2=16\cdot 36\end{array}\right\|\implies$ $\left\{\begin{array}{cc} b^2=26\cdot 36\\\\ c=6\sqrt{10}\end{array}\right\|\implies$ $\left\{\begin{array}{cc} b=6\sqrt{26}\cdot 36\\\\ c^2=10\cdot 36\end{array}\right\|\ .$

Proof 2. Let $K\in BC$ so that $AK\perp BC$ , i.e. $AB=AD\iff KB=KD=10$ . If denote $AK=x$ , then $KB=10$ , $KC=26$ and $\left\{\begin{array}{c} b^2=x^2+26^2\\\\ c^2=x^2+10^2\end{array}\right\|\implies$ $\left\{\begin{array}{cc} b^2+c^2=36\cdot 36\\\\ b^2-c^2=16\cdot 36\end{array}\right\|\implies$ $\left\{\begin{array}{cc} b^2=26\cdot 36\\\\ c=6\sqrt{10}\end{array}\right\|\implies$ $\left\{\begin{array}{cc} b=6\sqrt{26}\cdot 36\\\\ c^2=10\cdot 36\end{array}\right\|\ .$ Remark. You can use directly the equivalence $XY\perp AB\iff XA^2-XB^2=YA^2-YB^2$ .