Since $HG \parallel AB$ and $BG \parallel AH$, we have $BG \perp BC$ and $CH \perp GH$. Therefore, the quadrilateral $BGCH$ is cyclic. Since $H$ is the orthocenter of the triangle $ABC$, we have $\angle HAC = 90^\circ - \angle ACB = \angle CBH$. Using that $BGCH$ and $CGJI$ are cyclic quadrilaterals, we get
$$\angle CJI = \angle CGH = \angle CBH = \angle HAC.$$
Let $M$ be the intersection of $AC$ and $GH$, and let $D \neq A$ be the point on the line $AC$ such that $AH = HD$. Then $\angle MJI = \angle HAC = \angle MDH$.
Since $\angle MJI = \angle MDH$, $\angle IMJ = \angle HMD$, and $IM = MH$, the triangles $IMJ$ and $HMD$ are congruent, and thus $IJ = HD = AH$.

Obtain $\angle CGH = \angle HAC$ as in the previous solution. In the parallelogram $ABGH$ we have $\angle BAH = \angle HGB$. It follows that
$$\angle HMC = \angle BAC = \angle BAH + \angle HAC = \angle HGB + \angle CGH = \angle CGB.$$
So the right triangles $CMH$ and $CGB$ are similar. Also, in the circumcircle of triangle $GCI$ we have similar triangles $MIJ$ and $MCG$. Therefore,
$$\frac{IJ}{CG} = \frac{MI}{MC} = \frac{MH}{MC} = \frac{GB}{GC} = \frac{AH}{CG}.$$
Hence $IJ = AH$.