The line AC has the equation \(2y = x + 4\), which simplifies to \(y = \frac{1}{2}x + 2\). The gradient of AC is \(\frac{1}{2}\).

The line BD is perpendicular to AC, so its gradient is the negative reciprocal of \(\frac{1}{2}\), which is \(-2\).

The equation of line BD passing through \(D(10, -3)\) with gradient \(-2\) is:

\(y + 3 = -2(x - 10)\)

Simplifying gives:

\(y + 3 = -2x + 20\)

\(y = -2x + 17\)

Equating the equations of lines AC and BD:

\(\frac{1}{2}x + 2 = -2x + 17\)

Solving for \(x\):

\(\frac{1}{2}x + 2x = 17 - 2\)

\(\frac{5}{2}x = 15\)

\(x = 6\)

Substitute \(x = 6\) back into \(y = \frac{1}{2}x + 2\):

\(y = \frac{1}{2}(6) + 2 = 3 + 2 = 5\)

Thus, B is \((6, 5)\).

Since \(AB = BC\), the vector from A to B is the same as from B to C. If A is \((0, 2)\), then the vector AB is \((6 - 0, 5 - 2) = (6, 3)\).

Adding this vector to B gives C:

C = \((6 + 6, 5 + 3) = (12, 8)\).