It seems there is no need of $CF=BC$? Best regards, sunken rock

The condition is necessary. let $K$ be the midpoint of the arc $\overarc{BD}$ since $\angle BAD =90^\circ $ then $K$ is the circumcenter of $(BCD)$ hence $\angle CFB=\angle BAF+\angle ABF=45^\circ +\angle ABF $but $BFC$ is isosceles thus $\angle CFB=90^\circ -\frac 1 2\angle BCF=90^\circ -\frac 1 4\angle BKA=90^\circ -\frac 1 4\angle BDA=90^\circ -\frac 1 4(90^\circ -ABF)$ therefore we deduce $\angle ABF=30^\circ$ then $\angle ABC=105^\circ $ RH HAS

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Pls disregard my previous post (#2)! Clearly $C$ is the excenter of $\triangle ABD$ and, if $m(\widehat{ABD})=\alpha$, then $m(\widehat{CFD})=45^\circ+\alpha$ and $m(\widehat{CBF})=90^\circ-\frac{\alpha}2$ and from the equality of the last 2 angles, we get $\alpha=30^\circ$, a.s.o. Best regards, sunken rock