In triangle $CAT$, we have $\angle ACT = \angle ATC$ and $\angle CAT = 36^\circ$. If $\overline{TR}$ bisects $\angle ATC$, then $\angle CRT =$

[asy]
pair A,C,T,R;
C = (0,0); T = (2,0); A = (1,sqrt(5+sqrt(20))); R = (3/2 - sqrt(5)/2,1.175570);
draw(C--A--T--cycle);
draw(T--R);
label("$A$",A,N);
label("$T$",T,SE);
label("$C$",C,SW);
label("$R$",R,NW);
[/asy]

$\text{(A)}\ 36^\circ \qquad \text{(B)}\ 54^\circ \qquad \text{(C)}\ 72^\circ \qquad \text{(D)}\ 90^\circ \qquad \text{(E)}\ 108^\circ$
1. Given that $\triangle CAT$ is isosceles with $\angle ACT = \angle ATC$ and $\angle CAT = 36^\circ$, we can determine the measures of $\angle ACT$ and $\angle ATC$.
   
   Since the sum of the angles in any triangle is $180^\circ$, we have:
   \[
   \angle CAT + \angle ACT + \angle ATC = 180^\circ
   \]
   Given $\angle CAT = 36^\circ$ and $\angle ACT = \angle ATC$, let $\angle ACT = \angle ATC = x$. Then:
   \[
   36^\circ + x + x = 180^\circ
   \]
   Simplifying, we get:
   \[
   36^\circ + 2x = 180^\circ
   \]
   \[
   2x = 144^\circ
   \]
   \[
   x = 72^\circ
   \]
   Therefore, $\angle ACT = \angle ATC = 72^\circ$.

2. Since $\overline{TR}$ bisects $\angle ATC$, it divides $\angle ATC$ into two equal parts. Thus:
   \[
   \angle ATR = \angle RTC = \frac{72^\circ}{2} = 36^\circ
   \]

3. We need to find $\angle CRT$. Notice that $\angle CRT$ is an exterior angle for $\triangle ART$ at vertex $R$. The exterior angle theorem states that an exterior angle of a triangle is equal to the sum of the two non-adjacent interior angles. Therefore:
   \[
   \angle CRT = \angle CAT + \angle ATR
   \]
   Substituting the known values:
   \[
   \angle CRT = 36^\circ + 36^\circ = 72^\circ
   \]

The final answer is $\boxed{72^\circ}$