First, find the midpoint M of BD. The formula for the midpoint is:

\(M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)\)

Substitute the coordinates of B (2, 10) and D (6, 2):

\(M = \left( \frac{2 + 6}{2}, \frac{10 + 2}{2} \right) = (4, 6)\)

Next, find the gradient of BD:

\(m_{BD} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{2 - 10}{6 - 2} = -2\)

Since ABCD is a rhombus, the diagonals are perpendicular. Therefore, the gradient of AC is the negative reciprocal of \(m_{BD}\):

\(m_{AC} = \frac{1}{2}\)

The equation of line AC passing through M (4, 6) is:

\(y - 6 = \frac{1}{2}(x - 4)\)

Since A lies on the x-axis, \(y = 0\):

\(0 - 6 = \frac{1}{2}(x - 4)\)

\(-6 = \frac{1}{2}x - 2\)

\(-4 = \frac{1}{2}x\)

\(x = -8\)

Thus, A is (-8, 0).

To find C, use the vector method. The vector from A to M is:

\(\overrightarrow{AM} = (4 - (-8), 6 - 0) = (12, 6)\)

Since M is the midpoint, C is:

\(C = M + \overrightarrow{AM} = (4, 6) + (12, 6) = (16, 12)\)

(i) To find the equation of AC, use the points A(0, 4) and C(8, 0). The gradient \(m\) is given by \(m = \frac{0 - 4}{8 - 0} = -\frac{1}{2}\). The equation of the line is \(y = mx + c\). Substituting the point A(0, 4), we get \(4 = -\frac{1}{2}(0) + c\), so \(c = 4\). Therefore, the equation of AC is \(y = -\frac{1}{2}x + 4\).

For OB, since AC is the line of symmetry, OB is perpendicular to AC. The gradient of OB is the negative reciprocal of \(-\frac{1}{2}\), which is 2. Therefore, the equation of OB is \(y = 2x\).

(ii) To find the coordinates of B, use the midpoint formula for OB. Since O is the origin (0, 0), the midpoint of OB is \(\left( \frac{0 + x}{2}, \frac{0 + y}{2} \right)\). Given that the midpoint is on AC, substitute into the equation \(y = -\frac{1}{2}x + 4\). Solving the simultaneous equations \(y = 2x\) and \(y = -\frac{1}{2}x + 4\), we find \(x = 3.2\) and \(y = 6.4\). Therefore, the coordinates of B are (3.2, 6.4).

(i) To find the equation of BC, calculate the slope using points B(4, 6) and C(x, y):

\(m_{BC} = \frac{y - 6}{x - 4} = \frac{1}{2}\)

The equation of line BC is:

\(y - 6 = \frac{1}{2}(x - 4)\)

For CD, since BC is perpendicular to CD, \(m_{BC} \times m_{CD} = -1\).

\(m_{CD} = -2\)

Using point D(12, 5), the equation of line CD is:

\(y - 5 = -2(x - 12)\)

(ii) To find the coordinates of C, solve the simultaneous equations:

\(2y = x + 8\)

\(y + 2x = 29\)

Substitute \(x = 2y - 8\) into the second equation:

\(y + 2(2y - 8) = 29\)

\(y + 4y - 16 = 29\)

\(5y = 45\)

\(y = 9\)

Substitute \(y = 9\) back into \(x = 2y - 8\):

\(x = 2(9) - 8 = 10\)

Thus, the coordinates of C are (10, 9).

(i) The slope of AB is calculated as \(m = \frac{6 - 2}{1 - 3} = -2\). Since BC is perpendicular to AB, the slope of BC is \(\frac{1}{2}\). Using point B (1, 6), the equation of BC is \(y - 6 = \frac{1}{2}(x - 1)\) or \(2y = x + 11\).

(ii) To find the coordinates of C, solve the equations \(y = x - 1\) and \(y - 6 = \frac{1}{2}(x - 1)\). Substituting \(y = x - 1\) into the second equation gives \(x - 1 - 6 = \frac{1}{2}(x - 1)\). Solving this, we find \(x = 13\) and \(y = 12\), so C is (13, 12).

(iii) The length of AB is \(\sqrt{(3 - 1)^2 + (2 - 6)^2} = \sqrt{20}\). The length of BC is \(\sqrt{(13 - 1)^2 + (12 - 6)^2} = \sqrt{180}\). The perimeter is \(2 \times \sqrt{20} + 2 \times \sqrt{180} = 35.8\) or \(35.7\) or \(16\sqrt{5}\) or \(\sqrt{1280}\).

(i) To find the equation of AD, we first find the slope of AB which is \(m_{AB} = \frac{-3 - 6}{5 - 2} = -3\). Since AD is perpendicular to AB, the slope of AD is \(m_{AD} = \frac{1}{3}\). Using point A (2, 6), the equation of AD is \(y - 6 = \frac{1}{3}(x - 2)\) or \(3y = x + 16\).

(ii) To find the coordinates of D, we use the equation of CD: \(y - 3 = -3(x - 8)\) or \(y = -3x + 27\). Solving the equations of AD and CD simultaneously, we find \(D = (6\frac{1}{2}, 7\frac{1}{2})\).

(iii) To find the length of BE, we use the midpoint formula to find E such that ABCE is a parallelogram. The midpoint of AC is \((5, 4.5)\), so E is \((5, 12)\). The length of BE is \(\sqrt{(5 - 5)^2 + (12 + 3)^2} = 15\).