We let $D,E$ outside the triangle such that $B',A'$ are the midpoints of $CE,CD$. Then $\angle EAC=\angle DBC=90$. Also $A'B'\parallel DE$. Let $F$ be outside the triangle such that $AF=BF$ and $\angle AFB=180-\varphi$. Let $M$ be the midpoint of $DE$. Then it is well known that $\angle MAB=\angle MBA=\angle ECA$. Thus we have $\angle MAF=\angle MBF=90$. Also note that $\angle AFM=\angle ACE$, so $\triangle AFM\sim \triangle ACE$, and by spiral symmetry, $\triangle AME\sim \triangle AFC$. But since $EA\perp CA$, we have $EM\perp CF$, so $CF\perp A'B'$, and $F$ lies on the perpendicular bisector of $AB$, so $F\equiv C_1$, and $\angle AC_1B=180-\varphi$.

Your solution does not seem to be correct. $ \angle AC_{1}B=180-\varphi $ is not true as my initial graphic suggests. I will check and post my solution. It's a terrible allergy I'm having in Northern California.

Correct answer is $ \angle AC_{1}B=2\varphi $. Whoever "solved" it the first time got it wrong.

I think I know what the problem is. My definition of "base angle" is the angle opposite the base, but maybe the correct definition is the angle beside the base. That is probably how I got $180-\varphi$ but you got $2\varphi$. The solution should still be correct.