As $\triangle ADB \sim \triangle CAB$, 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 $\triangle AEB \sim \triangle CFB \Rightarrow \angle ABE = \angle CBF$.
As $EF$ is the midline of $\triangle CAD$, 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 $\angle ABM = \angle ABE + \angle EBM = 90^\circ, AC||BM$.