Problem 2

Because $MH\parallel AD$ and $NH\parallel AC$, we have $\angle HMN=\angle A'DC=\angle ADC=\beta$ and, using $E,T,M$ collinear, $\angle TMN=\angle CME=2\angle CAE=2\beta$. Thus $MH$ bisects $\angle NMT$. Symmetrically $\angle MNT=2\alpha$, and $NH$ bisects $\angle MNT$. It remains to choose the internal, not external, bisectors. Take $\triangle ACD$ counterclockwise. Since $C,M,N,D$ occur in this order and $C,D$ are the farther intersections with $MN$, the angle $\angle NHM=\angle CAD$ is obtuse. Reflecting the line $MN$ in the two bisectors $MH,NH$ gives lines meeting at $T$ on the same side of $MN$ as $H$; hence $H$ lies inside $\triangle TMN$, so these are its internal bisectors. Therefore $H$ is the incenter and $\angle NTH=\angle HTM=90^\circ-(\alpha+\beta)$. In $\triangle ADF$, the centre $N$ gives $\angle NFA=90^\circ-\angle ADF=90^\circ-(\alpha+\beta)$ (directed angles modulo $180^\circ$). Since $FA=AP$ and this equals $\angle NTH$, we obtain $HT\parallel AP$. Step 4 showed that the tangent to $(BEF)$ at $T$ is parallel to $AP$; the line through $H$ parallel to $AP$ therefore passes through $T$ and is exactly that tangent.