(H, E; B, C) is a harmonic bundle since <HEA = 90 and EA bisects <FEG. This means H is collinear with the feet of the altitudes from B and C in ABC, and it is also the orthocenter of AH’M where H’ and M are the orthocenter of ABC and the midpoint of BC. If Ha is the reflection of H’ over BC then J is the intersection of HHa with (ABC), so the reflection of J over BC is the foot from H’ onto AM, which is the reflection of the intersection of AM with (ABC) over M, so AJ is a symmedian.

Let $AK$ be $A-$symmedian where $K\in (ABC)$. Let $AD\cap (ABC)=A,M$ and $N$ be the midpoint of $BC$. Let $MN\cap AC=X$. We will show that $A,I,K,H$ are cyclic. Take the inversion centered at $A$ with radius $\sqrt{AB.AC}$ and reflect inverted points according to $AD$. $D\leftrightarrow M,B\leftrightarrow B,C\leftrightarrow C,K\leftrightarrow N$ And $I^*$ is the intersection of $AE$ with $(ABC)$. $F^*,M,G^*$ are collinear and $F^*G^*\perp AM$. $H^*$ is the miquel point of $F^*G^*BC$. Let's prove that $I^*,N,H^*$ are collinear. Since $H^*$ is the center of spiral homothethy mapping $BC$ to $G^*F^*$ and $N,M$ are midpoints, we have $H^*BNC\sim H^*G^*MF^*$. \[90-\angle XCH^*=90-\angle ACH^*=90-\angle AMH^*=\angle H^*MF^*=\angle H^*NC=90-\angle XNH^*\implies \angle XCH^*=\angle XNH^*\]Thus, $X,N,C,H^*$ are cyclic. \[\angle CH^*N=\angle CXN=90-\angle C=\angle CAI^*=\angle CH^*I^*\]Which gives that $H^*,N,I$ are collinear as desired.$\blacksquare$

Let $J'$ be point on $ABC$ such that $AJ'$ is symmedian. we will prove $J'$ is $J$. Let $M$ be midpoint of $BC$. Let $AE$ meet $ABC$ at $K$. Claim $1: H,K,J'$ are collinear. Proof $:$ Let $KJ'$ meet $BC$ at $H'$. Note that $\frac{H'B}{H'C}=\frac{BK}{CK}.\frac{BJ'}{CJ'} = \frac{\sin{90-B}}{\sin{90-C}}.\frac{\sin{C}}{\sin{B}}$. Now note that $HE$ is the external angle bisector of $\angle FEG$ so $\frac{HF}{HG} = \frac{EF}{EG} = \frac{\sin{90-B}}{\sin{90-C}}$. Also note that $HF = HB.\frac{\sin{B}}{\sin{AFG}}$ and $HG = HC.\frac{\sin{C}}{\sin{AGF}}$ so $\frac{HF}{HG} = \frac{HB}{HC}.\frac{\sin{B}}{\sin{C}}$ so $\frac{HB}{HC} = \frac{\sin{90-B}}{\sin{90-C}}.\frac{\sin{C}}{\sin{B}}$ which implies $H$ and $H'$ are same. Claim $2: AHKM$ is cyclic. Proof $:$ Note that $MEKJ'$ is cyclic so $\angle HKA = \angle EMJ' = \angle EMA = \angle HMA$ so $AHKM$ is cyclic. Note that $\angle J'AI = \angle MAK = \angle MHK = \angle IHJ'$ so $HJ'IA$ is cyclic which implies $J'$ is $J$ as wated.