BxMO Team Selection Test, March 2020

We will prove that $|FD|=|FC|$ and $|FC|=|FM|$, which proves the statement.

In the cyclic quadrilateral $ABCF$, we have $\angle BCF=180^{\circ }-\angle BAF$. As $|AB|=|BC|=|BD|$ we also have $\angle BAF=\angle BAD=\angle ADB$, hence $\angle BDF=180^{\circ }-\angle ADB=180^{\circ }-\angle BAF$. We see that $\angle BCF=\angle BDF$. Moreover, $\angle DFB=\angle AFB$ and $\angle CFB$ are inscribed angles on chords $AB$ and $BC$ of the same length, hence $\angle DFB=\angle CFB$. Triangles $BCF$ and $BDF$ have two pairs of equal angles; because they also have the side $BF$ in common, they are congruent. We conclude that $|FC|=|FD|$ and $\angle DBF=\angle CBF$.

The inscribed angle theorem yields $\angle CMF=2\angle CBF$. From the equality $\angle DBF=\angle CBF$ we just found, we get that $2\angle CBF=\angle CBD=60^{\circ }$, hence $\angle CMF=60^{\circ }$. Moreover, $|MC|=|MF|$ (radius of the circle), hence $△CMF$ is isosceles with an angle of $60^{\circ }$, which yields that the triangle is equilateral. This means that $|FC|=|FM|$.

This concludes the proof that $|FD|=|FC|=|FM|$.