67th Romanian Mathematical Olympiad

a) From $m(\angle BMA)=75^{\circ }$, it follows $m(\angle BAM)=180^{\circ }-75^{\circ }-45^{\circ }=60^{\circ }$, $m(\angle MAC)=30^{\circ }$ and $m(\angle MAF)=15^{\circ }$.

From $BF=AB$ it follows that $m(\angle AFB)=m(\angle BAF)=75^{\circ }$, hence $m(\angle ABF)=30^{\circ }$, whence, using $m(\angle BAM)=60^{\circ }$, it follows that $AM\perp BF$.

b) Take the point $D$ in the half-plane determined by $BC$ not containing $A$, so that $BD=BF$ and $m(\angle MBD)=15^{\circ }$.



Then the triangle $ABD$ is equilateral and the points $A$, $M$ and $D$ are collinear.

The triangle $ADC$ is isosceles, with $m(\angle DAC)=30^{\circ }$, hence $m(\angle ADC)=75^{\circ }=m(\angle CMD)$. Therefore the triangle $CDM$ is isosceles, hence $[CM]=[CD]$.

In the isosceles triangle $BDF$, the bisector $BC$ is the perpendicular bisector of the segment $[DF]$, hence the triangle $CFD$ is also isosceles, whence $[CF]=[CD]=[CM]$, that is the triangle $CFM$ is isosceles.