In triangle $ABC$, $\angle ABC$ is obtuse.  Point $D$ lies on side $AC$ such that $\angle ABD$ is right, and point $E$ lies on side $AC$ between $A$ and $D$ such that $BD$ bisects $\angle EBC$.  Find $CE$ given that $AC=35$, $BC=7$, and $BE=5$.
1. **Define the angles and segments:**
   Let $\alpha = \angle CBD = \angle EBD$ and $\beta = \angle BAC$. Also, let $x = CE$.

2. **Apply the Law of Sines in $\triangle ABE$:**
   \[
   \frac{\sin (90^\circ - \alpha)}{35 - x} = \frac{\sin \beta}{5}
   \]
   Since $\sin (90^\circ - \alpha) = \cos \alpha$, this equation becomes:
   \[
   \frac{\cos \alpha}{35 - x} = \frac{\sin \beta}{5}
   \]

3. **Apply the Law of Sines in $\triangle ABC$:**
   \[
   \frac{\sin (\alpha + 90^\circ)}{35} = \frac{\sin \beta}{7}
   \]
   Since $\sin (\alpha + 90^\circ) = \cos \alpha$, this equation becomes:
   \[
   \frac{\cos \alpha}{35} = \frac{\sin \beta}{7}
   \]

4. **Equate the expressions for $\frac{\cos \alpha}{\sin \beta}$:**
   From the first equation:
   \[
   \frac{\cos \alpha}{\sin \beta} = \frac{35 - x}{5}
   \]
   From the second equation:
   \[
   \frac{\cos \alpha}{\sin \beta} = \frac{35}{7}
   \]

5. **Set the two expressions equal to each other:**
   \[
   \frac{35 - x}{5} = \frac{35}{7}
   \]

6. **Solve for $x$:**
   \[
   \frac{35 - x}{5} = 5
   \]
   \[
   35 - x = 25
   \]
   \[
   x = 10
   \]

The final answer is $\boxed{10}$.