a) We have $\triangle AGK \sim \triangle AHL$ (1), because $\angle AGK = \angle AHL = 45^\circ$, and $\angle GAK = 90^\circ - \angle CAD = \angle HAL$. Thus, $\frac{AK}{AL} = \frac{AG}{AH}$. Since $AG = AE = AC$ and $AH = AB$, it follows that $\frac{AK}{AC} = \frac{AL}{AB}$, therefore $LK \parallel BC$.

b) From (1), we have $\angle AKG = \angle ALH$, so $ALNK$ is cyclic. Therefore, $\angle NAK = \angle NLK = \angle NIG = 45^\circ$, which shows that $AN$ is the bisector of the angle $BAC$. The bisector from $B$ in the triangle $ABC$ is parallel to the bisector of the angle $BAI$, which is also an altitude in the triangle $BAI$. Similarly, $CM$ is the external bisector from $C$ in the triangle $ABC$. Thus, $M$ is the center of the excircle relative to $A$ of the triangle $ABC$, so $M \in AN$.

Alternative solution for b).

Let $O = KL \cap AN$. Since $A, G, I$ are collinear and $DA \perp GI, DA \perp BC$, we get $GI \parallel BC \parallel KL$. Hence, $\frac{OK}{OL} = \frac{AG}{AI} = \frac{AG}{AH} = \frac{AK}{AL}$, therefore, by the converse of the angle bisector theorem, $AO$ is the bisector of the angle $LAK$. Let $P = AM \cap BC$. Then $\frac{CP}{PB} = \frac{AG}{AI} = \frac{AC}{AB}$, so $AP$ is the bisector of the angle $BAC$, hence both $M$ and $N$ lie on the angle bisector of the angle $BAC$, proving the collinearity of $A, M, N$.