Problem 1953

a) Let $E'$, $F'$ be the midpoints of $AE$, $BF$. Note that triangles $DFB$ and $CEA$ are right at $F$, $E$ and $I$ is the midpoint of $BD$ and $AC$ then triangles $IBF$, $ICE$ are isosceles at $I$. On the other hand, because $\angle GAI = \angle GBI$ then $\triangle IFB \sim \triangle IAE$, which implies
$$\angle IE'E = \frac{\angle AIE}{2} = \frac{\angle BIF}{2} = \angle F'IF$$
or $IE'$, $IF'$ are isogonal with respect to $\angle EIF$.

Therefore, we get
$$\begin{aligned} I(HE', FE) &= E(HE', FI) = E(F'G, FI) \\ &= I(F'G, FE) = I(GF', EF). \end{aligned}$$
Combining with $IE'$, $IF'$ are isogonal with respect to $\angle EIF$, we obtain that $IG$, $IH$ are isogonal with respect to $\angle EIF$, or $\angle AIB$. Hence, it suffices to show that if $AF$ meets $BE$ at $L$ then $IL$, $IG$ are isogonal with respect to $\angle AIB$. It is clear that
$$\triangle IAF \stackrel{\perp}{\sim} \triangle IEB,$$
then
$$(LA, LB) \equiv (IF, IB) \equiv (IA, IE) \pmod{\pi},$$
which means $L$ lies on $(IAE)$ and $(IBF)$. Let $G'$ be the intersection of $IL$ and the circumcircle of triangle $LAB$, we obtain that
$$\angle G'AB = \angle G'LB = \angle ILB = \angle IAE,$$
which implies that $AG$, $AG'$ are isogonal with respect to $\angle IAB$.
Similarly, we can point out that $BG'$, $BG$ are isogonal with respect to $\angle IBA$ then $G'$ is the isogonal conjugate of $G$ in triangle $IAB$. Hence, $IG$, $IL$ are isogonal with respect to $\angle AIB$.

b) Firstly, we will prove the following lemmas

Lemma 1. Denote $O$ to be the circumcenter of triangle $GAB$ then $IO, IL$ are isogonal with respect to $\angle AIB$.

*Proof*. The circumcircle of triangle $G'AB$ meets $IB, IA$ at $A_1, B_1$ respectively. By angle chasing, we have
$$\angle G'A_1I = \angle G'LB = \angle G'AB = \angle GAI.$$
Note that $A_1B_1$ and $AB$ are isogonal with respect to $\angle AIB$, then
$$\triangle IG'A_1 \sim \triangle IGA, \triangle IAB \sim \triangle IA_1B_1.$$
Because $O, J$ are the circumcenters of triangle $GAB, G'A_1B_1$ then
$$\triangle IAB \sim \triangle IA_1B_1,$$
in which $G, O$ are corresponding to $G', J$, implies that $IO,IJ$ are isogonal with respect to $\angle AIB$. $\square$

Lemma 2. $IJ \perp EF$ and the lines $IX, IJ$ are isogonal with respect to $\angle AIB$.

*Proof*. Let $O_1, O_2$ and $O_3$ be the circumcenters of triangles $LEF, LAE$ and $LAF$. It is clear that
$$\angle O_1O_2O_3 = \angle ILE = \angle IAE = \angle IBF = \angle ILE = \angle JO_2O_3$$
or $O_2O_3$ is the bisector of $\angle O_1O_2J$ and $LI$ is the bisector of $\angle ELF$. Similarly, $O_3O_2$ is the bisector $\angle O_1O_3J$, hence $O_1$ is the reflection of $J$ with respect to $O_2O_3$. On the other hand, because $LI$ is the common chord of $(O_2)$ and $(O_3)$ then $L$ is the reflection of $I$ with respect to $O_2O_3$. Therefore, $LIO_1J$ is an isosceles trapezoid, which means the reflection of $JI$ with respect to the bisector $LI$ of $\angle LEF$ passes through the circumcenter of triangle $ELF$, or $IJ \perp EF$.
Let $J'$ be the reflection of $J$ with respect to $EF$, it is clear that
$$\angle J'FE = \angle JFE = 90^\circ - \angle IJF = \angle NFE = \angle XFE$$
or $FJ'$, $FX$ are isogonal with respect to $\angle IFE$. Similarly, $EJ'$, $EX$ are isogonal with respect to $\angle IEF$, which means $IX$, $IJ'$ are isogonal with respect to $\angle EIF$ or $\angle AIB$. $\square$

Combining these two lemmas, we obtain that $IO$ passes through $X$ or $IO$, $EM$ and $FN$ are concurrent. To finish this problem, we need to point out that $IO$ passes through the circumcenter of triangle $KCD$.
Let $G'$ be the reflection of $G$ over $I$, then $G'C \parallel GA$ or $G'C \perp KC$. Similarly, $G'D \perp KD$ or $G'K$ is the diameter of ($KCD$). Therefore, the reflection of $G$ over $I$ lies on ($KCD$), or ($KCD$) and ($GAB$) are symmetric with respect to $I$, which means $IO$ passes through the circumcenter of triangle $KCD$. $\square$