In triangle $ABC, AB = AC = 10$ and $BC = 12$. Point $D$ lies strictly between $A$ and $B$ on $\overline{AB}$ and point $E$ lies strictly between $A$ and $C$ on $\overline{AC}$ so that $AD = DE = EC$.  Find $AD.$
By the Law of Cosines on triangle $ABC,$
\[\cos A = \frac{10^2 + 10^2 - 12^2}{2 \cdot 10 \cdot 10} = \frac{7}{25}.\]Let $x = AD = DE = CE.$

[asy]
unitsize(0.5 cm);

pair A, B, C, D, E;
real x = 250/39;

A = (0,8);
B = (-6,0);
C = (6,0);
D = interp(A,B,x/10);
E = interp(A,C,(10 - x)/10);

draw(A--B--C--cycle);
draw(D--E);

label("$A$", A, N);
label("$B$", B, SW);
label("$C$", C, SE);;
label("$D$", D, NW);
label("$E$", E, NE);

label("$x$", (A + D)/2, NW);
label("$x$", (D + E)/2, SE);
label("$x$", (C + E)/2, NE);
label("$10 - x$", (A + E)/2, NE);
[/asy]

Then by the Law of Cosines on Triangle $ADE$,
\[x^2 = x^2 + (10 - x)^2 - 2x(10 - x) \cos A = x^2 + (10 - x)^2 - 2x(10 - x) \cdot \frac{7}{25}.\]Then
\[(10 - x)^2 - 2x(10 - x) \cdot \frac{7}{25} = 0.\]Since $x \neq 10,$ we can divide both sides by $10 - x,$ to get
\[10 - x - 2x \cdot \frac{7}{25} = 0.\]Solving, we find $x = \boxed{\frac{250}{39}}.$