Solution:
Let $L$ be the foot of the perpendicular from $G$ to $EC$ and let $Q$ be the point of intersection of the lines $EG$ and $BC$. Then,
$$(GCE) = \frac{1}{2} EC \cdot GL = \frac{1}{2} \sqrt{a^{2} + b^{2}} \cdot GL$$
So, $GL = \frac{a}{b} \sqrt{a^{2} + b^{2}}$.

Observing that the triangles $QCE$ and $ELG$ are similar, we have $\frac{a}{b} = \frac{GL}{EL}$, which implies that $EL = \sqrt{a^{2} + b^{2}}$, or in other words $L \equiv C$.

Consider the circumcircle $\omega$ of the triangle $EBC$. Since
$$\angle EBG = \angle ECG = \angle EHG = 90^{\circ}$$
the points $H$ and $G$ lie on $\omega$.

From the given similarity of the triangles $ACE$ and $AEI$, we have that
$$\angle AIE = \angle AEC = 90^{\circ} + \angle GEC = 90^{\circ} + \angle GHC = \angle EHC$$
therefore $EHCI$ is cyclic, thus $I$ lies on $\omega$.

Since $EB = EC$, we get that $\angle EIC = \angle EHB$, thus $\angle JIE = \angle EHK$. We conclude that $JIH K$ is cyclic, therefore
$$\angle JKH = \angle HIC = \angle HBC$$
It follows that $KJ \parallel BC$, so $KJ \perp AB$.