Solutions to this question by accurate drawing will not be accepted. The diagram, which is not drawn to scale, shows a quadrilateral $ABCD$ with vertices $A(-4, -6)$, $B(-7, h)$ and $C(2, 6)$ and $D$, which lies on the $x$-axis. $AD$ is parallel to $BC$.
The point $E$ lies on $AC$ such that $AE : AC = 1 : 3$. The line $BE$ produced meets the line $AD$ at $F$.

1. Express the squares of the lengths $AB$ and $BC$ in terms of $h$.

$AB^2 = (-7 - (-4))^2 + (h - (-6))^2 = (-3)^2 + (h + 6)^2 = 9 + (h + 6)^2$
$BC^2 = (2 - (-7))^2 + (6 - h)^2 = 9^2 + (6 - h)^2 = 81 + (6 - h)^2$

2. Equate $AB^2$ and $BC^2$ to solve for $h$.

Since $AB = BC$, we have $AB^2 = BC^2$:
$9 + (h + 6)^2 = 81 + (6 - h)^2$
$9 + h^2 + 12h + 36 = 81 + 36 - 12h + h^2$
$12h + 45 = -12h + 117$
$24h = 72$
$h = 3$ (shown)

1. Calculate the gradient of the line segment $AB$

Using $A(-4, -6)$ and $B(-7, 3)$:
$m_{AB} = \frac{3 - (-6)}{-7 - (-4)} = \frac{9}{-3} = -3$

2. Calculate the gradient of the line segment $BC$

Using $B(-7, 3)$ and $C(2, 6)$:
$m_{BC} = \frac{6 - 3}{2 - (-7)} = \frac{3}{9} = \frac{1}{3}$

3. Show that the product of the gradients is $-1$

$m_{AB} \times m_{BC} = -3 \times \frac{1}{3} = -1$
Since the product of their gradients is $-1$, $AB$ is perpendicular to $BC$. Therefore, $\angle ABC = 90^\circ$.

1. Calculate the gradient of the line segment $BC$

Gradient of $BC = \frac{6 - 3}{2 - (-7)} = \frac{3}{9} = \frac{1}{3}$

2. Use the parallel condition to find the coordinates of $D$

Since $D$ lies on the $x$-axis, let $D$ be $(x, 0)$.
Since $AD$ is parallel to $BC$, their gradients are equal.
Gradient of $AD = \frac{0 - (-6)}{x - (-4)} = \frac{6}{x + 4}$
$\frac{6}{x + 4} = \frac{1}{3}$
$x + 4 = 18$
$x = 14$
Coordinates of $D$ are $(14, 0)$.

1. Find the $x$-coordinate of $E$ using the given ratio

Since $AE : AC = 1 : 3$, the horizontal distance from $A$ to $E$ is $\frac{1}{3}$ of the horizontal distance from $A$ to $C$.
$x_E = -4 + \frac{1}{3}(2 - (-4))$
$x_E = -4 + \frac{1}{3}(6) = -2$

2. Find the $y$-coordinate of $E$ using the given ratio

Similarly, the vertical distance from $A$ to $E$ is $\frac{1}{3}$ of the vertical distance from $A$ to $C$.
$y_E = -6 + \frac{1}{3}(6 - (-6))$
$y_E = -6 + \frac{1}{3}(12) = -2$
The coordinates of $E$ are $(-2, -2)$.

1. Find the midpoint of the diagonal $AC$

The diagonals of a square bisect each other, so the midpoint of $AC$ is the same as the midpoint of $BG$.
Midpoint of $AC = \left( \frac{-4 + 2}{2}, \frac{-6 + 6}{2} \right) = (-1, 0)$

2. Use the midpoint to find the coordinates of $G$

Let the coordinates of $G$ be $(x, y)$.
Midpoint of $BG = \left( \frac{-7 + x}{2}, \frac{3 + y}{2} \right)$
Equating the midpoints:
$\frac{-7 + x}{2} = -1 \implies -7 + x = -2 \implies x = 5$
$\frac{3 + y}{2} = 0 \implies 3 + y = 0 \implies y = -3$
The coordinates of $G$ are $(5, -3)$.

1. Calculate the square of the length of $AB$

From earlier parts, $A(-4, -6)$ and $B(-7, 3)$.
$AB^2 = (-7 - (-4))^2 + (3 - (-6))^2$
$AB^2 = (-3)^2 + 9^2 = 9 + 81 = 90$

2. Calculate the area of triangle $ABC$ using the fact that it is a right-angled isosceles triangle

Since $\angle ABC = 90^\circ$ and $AB = BC$, triangle $ABC$ is a right-angled triangle.
Area $= \frac{1}{2} \times AB \times BC = \frac{1}{2} AB^2$
Area $= \frac{1}{2} \times 90 = 45$