$\angle BEC= \angle BDC = \tfrac{1}{2} \pi$. Let $AE=a, EC=b, CD=c$. Then $c^2+1=257 \implies c=16$ by Pythagoras. Now again, $AD^2+c^2 = (a+b)^2 \implies 144+256=(a+b)^2 \implies a+b=20$. By Power of a Point, $a(a+b)=12 \cdot 13 \implies a= \frac{3}{5} \cdot 13 = \frac{39}{5} \implies b= EC=\frac{61}{5}$ Now, draw altitude $AX$ to $BC$ and let $BX=d, CX= \sqrt{257}-d$ and apply Pythagoras to get $AX= \frac{13 \cdot 16}{\sqrt{257}}$.

Clearly $\angle{BDC} = 90^\circ$. By the Pythagorean Theorem, we see that $BD = 16$. Since $\angle{BDC} = 90^\circ$, this implies that $\angle{CDA} = 90^\circ$. By Pythagorean Theorem, $AC = 20$. We let $AE = r$, which implies $EC = 20 - r$. By power of a point, $156 = 20r$, $\implies r = \frac{39}{5}$. $\implies EC = \boxed{\frac{61}{5}}$. Let $AY$ be the altitude to $BC$. By the pythagorean theorem, we can build a system of equations. Let $BY = q$. $$\implies 169 - q^2 = AX^2$$$$\implies 143 + 2q\sqrt{257} = AX^2$$ Subtract the first equation from the seconds equation to get that: $$\implies 26 - 2d\sqrt{257} = 0$$$$\implies 2q\sqrt{257} = 26$$$$\implies q = \frac{13}{\sqrt{257}}$$$$\implies AX^2 = \frac{169(257 - 1)}{257} = \frac{169\cdot256}{257}$$$$\implies AX = \boxed{\frac{13\cdot16}{257}}$$