For a): $\angle{TNO}=\angle{TAO}=\angle{TMO}$ but $\angle{ONM}=\angle{OMN}$ so $TN=TM$ but $OM=ON$ $\implies TO\perp MN \implies MN\parallel BC \implies BM=CN \implies \angle{BAM}=\angle{CAN}$

For b) * Inversion with center at D, radius |DI| shows that it is equivalent to prove that the circumcircle of (APQ) touches circumcircle of (ABC) at A. * By Pascal's thm Q,I,P are collinear * Let ${R}=DF\cap BC, {S}=DE\cap BC$, it is easy to show that triangles (SIR) and (BAC) are homothetic with center at D, then PQ is parallel to BC * The rest is angle chasing.

My solutions

Any other solution

b) Since $\angle PFI=\angle DAN =\angle MAD=\angle PAI \Longrightarrow P,I,A,F$ and similarly $Q,I,A,E$ are concyclic $\Longrightarrow \angle DHI=\angle PID=\angle DFA \Longrightarrow IH\parallel AF, IK\parallel AE, HK\parallel FE \Longrightarrow \dfrac{DH}{DF}=\dfrac{DI}{DA}=\dfrac{DG}{DI} \Longrightarrow GH\parallel IF, GK\parallel IE$. The rest is simple angle chasing using the fact that $MN\parallel BC$.

Here is my solution for this problem Solution a) Let $S$ be a point which satisfies $\triangle OTM$ $\stackrel{+}{=}$ $\triangle OSN$ Then: $\triangle OTS$ $\stackrel{+}{\sim}$ $\triangle OMN$ So: $\triangle OST$ is $O$ - isosceles Combine with: $(TO; TM) \equiv (TN; TO) \equiv (SO; SN)$ (mod $\pi$), we have: $\triangle NTS$ is $N$ - isosceles Then: $TM = TN = NS$ or $\triangle TMO$ $\stackrel{-}{=}$ $\triangle TNO$ So: $(ON; OT) \equiv (OT; OM)$ (mod $\pi$) or $MN$ $\parallel$ $BC$ Hence: $AM$, $AN$ are isogonal conjugate with respect to $\triangle ABC$ b) We have: $(FD; FN) \equiv (AM; AD) \equiv (AD; AN) \equiv (EM; ED)$ (mod $\pi$) Then: $A$, $F$, $P$, $I$ lie on a circle; $A$, $E$, $Q$, $I$ lie on a circle So: $(IQ; IP) \equiv (IQ; IA) + (IA; IP) \equiv (ED; EA) + (FA; FD) \equiv 0$ (mod $\pi$) and $(IA; IP) \equiv (FA; FD) \equiv (BA; BD) \equiv (GA; GB)$ (mod $\pi$) or $\overline{P; I; Q}$ $\parallel$ $BC$ $\parallel$ $MN$ But: $ID^2 = \overline{DG} . \overline{DA} = \overline{DH} . \overline{DP} = \overline{DK} . \overline{DQ}$ then: 3 sets of points $(A; G; H; P)$, $(A; K; Q; G)$, $(K; Q; H; P)$ lie on a circle Hence: $(GH; GB) \equiv (GH; GA) + (GA; GB) \equiv (PD; PA) + (IA; IP) \equiv (PF; PA) + (FA; FP) \equiv (AF; AM)$ $\equiv (AF; AD) + (AD; AM) \equiv (PF; PI) + (AN; AD) \equiv (PD; PI) + (KD; KG) \equiv (KH; KQ) + (KQ; KG)$ $\equiv (KH; KG)$ (mod $\pi$) or $(KHG)$ tangents $BC$ at $G$