62nd Ukrainian National Mathematical Olympiad, Third Round, Second Tour

As $△ADB\sim △CAB$, we get $\frac{AD}{AB}=\frac{CA}{CB}$. As $AE=\frac{1}{2}AD$, $CF=\frac{1}{2}CA$, we get $\frac{AF}{AB}=\frac{CF}{CB}$. From this similarity we get that (fig. 16): $$\angle BAE=\angle BAD=\angle BCA=\angle BCF$$ Then by angle and the ratio of the sides we get that $△AEB\sim △CFB\Rightarrow \angle ABE=\angle CBF$. As $EF$ is the midline of $△CAD$, we also get $EF||BC\Rightarrow \angle BFE=\angle CBF$. Next, we get the following equalities of angles: $\angle BFE=\frac{1}{2}\angle BME=90^{\circ }-\angle EBM\Rightarrow \angle ABE=\angle CBF=90^{\circ }-\angle EBM$. Therefore $\angle ABM=\angle ABE+\angle EBM=90^{\circ },AC||BM$.