Altitude to altitude

Since $D$ is the midpoint of base $AB$ of the isosceles triangle, we have $BD = AD = \frac{13}{2}$ . And since $E$ lies on a circle with diameter $AB$ , $DE = DB = \frac{13}{2} = \boxed { 6.5 }$ .

A pure bashing solution:

Let $A=(0,0)$ . Then $D=(6.5,0)$ . Also, $CD=\sqrt{14^2-6.5^2}=\sqrt{153.75}$ . Therefore, $C=(6.5,\sqrt{153.75})$ .

Now, we can find the equation of CB, which is $y=-\frac{\sqrt{153.75}}{6.5}x+2\sqrt{153.75}$

Since AE is perpendicular to CB, we know that the product of their gradients is -1.

Hence, we can also find the equation of AE, which is $y=\frac{6.5}{\sqrt{153.75}}x$

We wish to find the coordinates of E, so we equate the two equations. $\frac{6.5}{\sqrt{153.75}}x=-\frac{\sqrt{153.75}}{6.5}x+2\sqrt{153.75}$

Solving, we get $E=\left(\frac{7995}{784},\frac{7995}{784}\times\frac{6.5}{\sqrt{153.75}}\right)$

Now, we use the distance formula to find DE. $DE=\sqrt{\left(\frac{7995}{784}-6.5\right)^2+\left(\frac{7995}{784}\times\frac{6.5}{\sqrt{153.75}}\right)^2}=6.5$

I just found it out by using analytical approach.