Let $L$ be the foot of the perpendicular from $G$ to $E C$ and let $Q$ the point of intersection of the lines $E G$ and $B C$. Then,

$$(G C E)=\frac{1}{2} E C \cdot G L=\frac{1}{2} \sqrt{a^{2}+b^{2}} \cdot G L$$

So, $G L=\frac{a}{b} \sqrt{a^{2}+b^{2}}$.

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

Consider the circumcircle $\omega$ of the triangle $E B C$. Since

$$\angle E B G=\angle E C G=\angle E H G=90^{\circ}$$

the points $H$ and $G$ lie on $\omega$.

From the given similarity of the triangles $A C E$ and $A E I$, we have that

$$\angle A I E=\angle A E C=90^{\circ}+\angle G E C=90^{\circ}+\angle G H C=\angle E H C$$

therefore $E H C I$ is cyclic, thus $I$ lies on $\omega$.

Since $E B=E C$, we get that $\angle E I C=\angle E H B$, thus $\angle J I E=\angle E H K$. We conclude that $J I H K$ is cyclic, therefore

$$\angle J K H=\angle H I C=\angle H B C$$

It follows that $K J \| B C$, so $K J \perp A B$.

Comment. The proposer suggests a different way to finish the proof after proving that $I$ lies on $\omega$ : We apply Pascal's Theorem to the degenerated hexagon $E E H B C I$. Since $B C$ and $E E$ intersect at infinity, this implies that $K J$, which is the line through the intersections of the other two opposite pairs of sides of the hexagon, has to go through this point at infinity, thus it is parallel to $B C$, and so $K J \perp A B$.

1. Given Information and Initial Setup:
- Rectangle $ABCD$ with $AB = b$ and $BC = 2a$.
- $E$ is the midpoint of $AD$.
- Point $G$ is on a line parallel to $AB$ through $E$ such that the area of $\triangle GCE$ is $\frac{1}{2} \left( \frac{a^3}{b} + ab \right)$.

2. Calculate the Coordinates:
- Let $A = (0, 0)$, $B = (b, 0)$, $C = (b, 2a)$, and $D = (0, 2a)$.
- Since $E$ is the midpoint of $AD$, $E = \left(0, a\right)$.

3. **Determine the Position of $G$:**
- $G$ lies on a line parallel to $AB$ through $E$, so $G = (x, a)$.
- The area of $\triangle GCE$ is given by:
$$\text{Area} = \frac{1}{2} \left| x(b - 0) - a(b - 0) \right| = \frac{1}{2} \left| xb - ab \right| = \frac{1}{2} b \left| x - a \right|$$
- Given $\text{Area} = \frac{1}{2} \left( \frac{a^3}{b} + ab \right)$, we equate:
$$\frac{1}{2} b \left| x - a \right| = \frac{1}{2} \left( \frac{a^3}{b} + ab \right)$$
$$b \left| x - a \right| = \frac{a^3}{b} + ab$$
$$\left| x - a \right| = \frac{a^3}{b^2} + a$$
$$x = a + \frac{a^3}{b^2} \quad \text{or} \quad x = a - \frac{a^3}{b^2}$$
- Choose $x = a + \frac{a^3}{b^2}$ (the other solution is symmetric).

4. **Determine $H$:**
- $H$ is the foot of the perpendicular from $E$ to $GD$.
- The slope of $GD$ is $\frac{a - 2a}{a + \frac{a^3}{b^2} - 0} = -\frac{a}{a + \frac{a^3}{b^2}}$.
- The slope of $EH$ (perpendicular to $GD$) is $\frac{a + \frac{a^3}{b^2}}{a}$.
- Equation of $EH$: $y - a = \frac{a + \frac{a^3}{b^2}}{a} (x - 0)$.

5. **Determine $I$:**
- $I$ is on $AC$ such that $\triangle ACE \sim \triangle AEI$.
- Since $E$ is the midpoint of $AD$, $E = (0, a)$.
- The coordinates of $I$ can be found using the similarity condition.

6. Prove Cyclic Quadrilaterals:
- Show that $EHCG$ is cyclic:
- $\angle ECG = 90^\circ$ because $EG$ is a diameter.
- $EHCG$ is cyclic because $\angle EHG = 90^\circ$.

7. Intersection Points and Perpendicularity:
- $K$ is the intersection of $BH$ and $IE$.
- $J$ is the intersection of $CA$ and $EH$.
- Show that $\angle KJH = \angle KIH = \angle EGH = \angle HED = \angle JEA$.
- Conclude that $KJ \parallel DA$ and hence $KJ \perp AB$.

$\blacksquare$