75th Romanian Mathematical Olympiad

a) We have $△AGK\sim △AHL$ (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 $A,M,N$.