The diagram shows a trapezium $ABCD$ in which $AD$ is parallel to $BC$ and angle $ABC = 90^\circ$.
The coordinates of $A$, $C$ and $D$ are $(0, 4)$, $(5, 3)$ and $(6, -5)$ respectively.

1. Find the gradient of the line $AD$ and hence the gradient of $BC$.

Gradient of $AD = \frac{-5 - 4}{6 - 0} = \frac{-9}{6} = -\frac{3}{2}$
Since $AD$ is parallel to $BC$, gradient of $BC = -\frac{3}{2}$.

2. Find the equation of the line $BC$.

Using point $C(5, 3)$ and gradient $-\frac{3}{2}$:
$y - 3 = -\frac{3}{2}(x - 5)$
$2y - 6 = -3x + 15$
$3x + 2y = 21$ --- (1)

3. Find the equation of the line $AB$.

Since $\angle ABC = 90^\circ$, $AB$ is perpendicular to $BC$.
Gradient of $AB = \frac{-1}{-3/2} = \frac{2}{3}$
Using point $A(0, 4)$ and gradient $\frac{2}{3}$:
$y - 4 = \frac{2}{3}(x - 0)$
$y = \frac{2}{3}x + 4$
$2x - 3y = -12$ --- (2)

4. Solve the equations simultaneously to find the coordinates of $B$.

From (2), $x = \frac{3y - 12}{2}$. Substitute into (1):
$3\left(\frac{3y - 12}{2}\right) + 2y = 21$
$9y - 36 + 4y = 42$
$13y = 78 \implies y = 6$
Substitute $y = 6$ into $x = \frac{3(6) - 12}{2} = \frac{6}{2} = 3$.
Therefore, the coordinates of $B$ are $(3, 6)$ (shown).

1. Relate the ratio of areas to the ratio of base lengths.

Triangles $BED$ and $BAD$ share the same perpendicular height from $B$ to the line $ED$.
Therefore, the ratio of their areas is equal to the ratio of their bases:
$\frac{ED}{AD} = \frac{4}{3}$

2. Find the coordinates of $E$ using vector translation or ratio.

Since $E$ lies on $DA$ produced, the points are in the order $D, A, E$.
This means $ED = EA + AD$, so $\frac{EA + AD}{AD} = \frac{4}{3} \implies \frac{EA}{AD} = \frac{1}{3}$.
Using vectors, $\overrightarrow{AE} = \frac{1}{3}\overrightarrow{DA}$.
$\overrightarrow{DA} = A - D = (0 - 6, 4 - (-5)) = (-6, 9)$
$\overrightarrow{AE} = \frac{1}{3}(-6, 9) = (-2, 3)$
Coordinates of $E = A + \overrightarrow{AE} = (0, 4) + (-2, 3) = (-2, 7)$.

1. Find the coordinates of $E$ using the ratio of areas from part (ii).

Since $\Delta BED$ and $\Delta BAD$ share the same height from $B$ to the line $ED$, the ratio of their areas is the ratio of their bases:
$\frac{ED}{AD} = \frac{4}{3}$
Since $E$ lies on $DA$ produced, $A$ is between $E$ and $D$, so $EA = \frac{1}{3}AD$.
$\vec{DA} = A - D = (0 - 6, 4 - (-5)) = (-6, 9)$
$\vec{AE} = \frac{1}{3}\vec{DA} = (-2, 3)$
$E = A + \vec{AE} = (0, 4) + (-2, 3) = (-2, 7)$

2. Determine the relationship between the diagonals of the kite $ECDF$.

In kite $ECDF$, the diagonals are $ED$ and $CF$. The diagonals of a kite are perpendicular, and one bisects the other.
Let's check the lengths of $EC$ and $CD$:
$EC^2 = (5 - (-2))^2 + (3 - 7)^2 = 49 + 16 = 65$
$CD^2 = (6 - 5)^2 + (-5 - 3)^2 = 1 + 64 = 65$
Since $EC = CD$, for $F$ to be uniquely determined as a kite (specifically a rhombus in this orientation), the diagonal $ED$ must bisect $CF$. Thus, the intersection of the diagonals is the midpoint of both $ED$ and $CF$.

3. Find the coordinates of $F$.

Midpoint of $ED$, $M = \left(\frac{-2 + 6}{2}, \frac{7 + (-5)}{2}\right) = (2, 1)$.
Since $M$ is also the midpoint of $CF$:
$\frac{x_F + 5}{2} = 2 \implies x_F = -1$
$\frac{y_F + 3}{2} = 1 \implies y_F = -1$
So, $F = (-1, -1)$.

4. Calculate the area of the kite $ECDF$.

Area of kite $= \frac{1}{2} \times ED \times CF$
$ED = \sqrt{(6 - (-2))^2 + (-5 - 7)^2} = \sqrt{8^2 + (-12)^2} = \sqrt{208}$
$CF = \sqrt{(5 - (-1))^2 + (3 - (-1))^2} = \sqrt{6^2 + 4^2} = \sqrt{52}$
Area $= \frac{1}{2} \times \sqrt{208} \times \sqrt{52} = \frac{1}{2} \times \sqrt{10816} = \frac{1}{2} \times 104 = 52$