Let $ABC$ be a triangle with $\angle BAC=117^\circ$.  The angle bisector of $\angle ABC$ intersects side $AC$ at $D$.  Suppose $\triangle ABD\sim\triangle ACB$.  Compute the measure of $\angle ABC$, in degrees.
1. Given that $\triangle ABD \sim \triangle ACB$, we know that corresponding angles are equal. Therefore, $\angle ABD = \angle ACB$.
2. Let $\angle ABD = x$. Since $\triangle ABD \sim \triangle ACB$, we also have $\angle ADB = \angle ABC$ and $\angle BAD = \angle BAC = 117^\circ$.
3. In $\triangle ABD$, the sum of the angles is $180^\circ$. Therefore, we have:
   \[
   \angle BAD + \angle ABD + \angle ADB = 180^\circ
   \]
   Substituting the known angles, we get:
   \[
   117^\circ + x + \angle ADB = 180^\circ
   \]
   Since $\angle ADB = \angle ABC$, we can rewrite the equation as:
   \[
   117^\circ + x + \angle ABC = 180^\circ
   \]
   Solving for $\angle ABC$, we get:
   \[
   \angle ABC = 180^\circ - 117^\circ - x = 63^\circ - x
   \]
4. Since $\triangle ABD \sim \triangle ACB$, we also have $\angle ACB = x$. In $\triangle ACB$, the sum of the angles is $180^\circ$. Therefore, we have:
   \[
   \angle BAC + \angle ABC + \angle ACB = 180^\circ
   \]
   Substituting the known angles, we get:
   \[
   117^\circ + \angle ABC + x = 180^\circ
   \]
   Solving for $\angle ABC$, we get:
   \[
   \angle ABC = 180^\circ - 117^\circ - x = 63^\circ - x
   \]
5. Since $\angle ABC = 2x$ (from the similarity condition $\triangle ABD \sim \triangle ACB$), we can set up the equation:
   \[
   2x = 63^\circ - x
   \]
   Solving for $x$, we get:
   \[
   3x = 63^\circ
   \]
   \[
   x = 21^\circ
   \]
6. Therefore, $\angle ABC = 2x = 2 \times 21^\circ = 42^\circ$.

The final answer is $\boxed{42^\circ}$.