Let $ABC$ be a non-isosceles triangle. Two isosceles triangles $AB'C$ with base $AC$ and $CA'B$ with base $BC$ are constructed outside of triangle $ABC$. Both triangles have the same base angle $\varphi$. Let $C_1$ be a point of intersection of the perpendicular from $C$ to $A'B'$ and the perpendicular bisector of the segment $AB$. Determine the value of $\angle AC_1B.$
1. **Construct Isosceles Triangles**:
   Let $D$ and $E$ be points outside the triangle $ABC$ such that $B'$ and $A'$ are the midpoints of $CE$ and $CD$, respectively. This construction ensures that $\triangle AB'C$ and $\triangle CA'B$ are isosceles with base angles $\varphi$.

2. **Angles in the Construction**:
   Since $B'$ and $A'$ are midpoints, we have:
   \[
   \angle EAC = \angle DBC = 90^\circ
   \]
   Also, $A'B' \parallel DE$ because $A'B'$ is the midline of $\triangle CDE$.

3. **Construct Point $F$**:
   Let $F$ be a point outside the triangle such that $AF = BF$ and $\angle AFB = 180^\circ - \varphi$. This construction ensures that $F$ is equidistant from $A$ and $B$.

4. **Midpoint $M$**:
   Let $M$ be the midpoint of $DE$. Since $A'B' \parallel DE$, $M$ lies on the perpendicular bisector of $DE$.

5. **Angle Relationships**:
   It is known that:
   \[
   \angle MAB = \angle MBA = \angle ECA
   \]
   This implies:
   \[
   \angle MAF = \angle MBF = 90^\circ
   \]

6. **Similarity of Triangles**:
   Note that $\angle AFM = \angle ACE$. Therefore, $\triangle AFM \sim \triangle ACE$ by AA similarity. By spiral symmetry, $\triangle AME \sim \triangle AFC$.

7. **Perpendicularity and Symmetry**:
   Since $EA \perp CA$, we have $EM \perp CF$. Thus, $CF \perp A'B'$. Also, $F$ lies on the perpendicular bisector of $AB$, so $F \equiv C_1$.

8. **Conclusion**:
   Therefore, $\angle AC_1B = 180^\circ - \varphi$.

The final answer is $\boxed{180^\circ - \varphi}$