Given $A(0,0),B(b,0),C(0,b)$. Given $\angle ABC=45^{\circ}$ and $\angle AMB=75^{\circ}\ ,\ M \in BC$, then $\angle BAM=60^{\circ}$. Equation of $AM\ :\ y=\sqrt{3}x$, equation of $BC\ :\ y=-x+b$, point $M(\frac{b}{\sqrt{3}+1},\frac{b\sqrt{3}}{\sqrt{3}+1})$. The line $AF\ :\ y=\tan 75^{\circ} \cdot x$ cuts the circle $(x-b)^{2}+y^{2}=b^{2}$ in the point $F(b(1-\frac{\sqrt{3}}{2},\frac{b}{2})$. Slope of the line $BF\ :\ m_{BF}=-\frac{1}{\sqrt{3}}$, slope of the line $AM\ :\ m_{AM}=\sqrt{3}$, so $BF \bot AM$. $\triangle CFM$ isosceles, $CM=CF$, $CM^{2}=CF^{2}=2-\sqrt{3}$.

Easily $\triangle ABF$ is $30^\circ-75^\circ-75^\circ$ and $\angle BAM=60^\circ$, thus $AM\bot BF$. Easily $ABMF$ is cyclic, thence $\angle FMC=\angle BAF=75^\circ$. Let $D=AM\cap BF$; from $\triangle ABD$ we get $AD=\frac{AB}2$, hence if $E$ is projection of $F$ onto $AC$, then $E$ is midpoint of $AC$ and $\angle ACF=\angle CAF=15^\circ$, implying $\angle BCF=30^\circ$ and $\triangle CFM$ is indeed isosceles. Best regards, sunken rock