To solve the problem, let's break it down into parts:

### a) Proving that the points \( M, A, O, B \) are concyclic.

In this situation, \( MA \) and \( MB \) are tangents drawn from point \( M \) to the circle, meaning that the angles \( \angle MAB \) and \( \angle MBA \) are equal to \( 90^\circ \) because the radius at the point of tangency is perpendicular to the tangent line.

In triangle \( MAB \):
- Since \( OA \) and \( OB \) are radii of the circle, we have \( OA = OB \).
- We thus have two angles \( \angle MAB \) and \( \angle MBA \) equal to \( 90^\circ \).

From this, \( \angle AOB = 180^\circ - (\angle MAB + \angle MBA) = 180^\circ - 90^\circ - 90^\circ = 0^\circ \), which shows that the points \( M, A, O, B \) lie on the same circle (the circumcircle of triangle \( MAB \)).

### b) Proving that \( MB \cdot BC = BK \cdot MO \).

By the Power of a Point theorem:
\[
MO^2 = MA \cdot MB
\]

We can express \( MO \) in terms of \( BK \) and the segments on the circle. Since \( BK \perp AC \) and \( AC \) is a diameter, triangles will exhibit the relationships of similar properties leading to \( MB \cdot BC = BK \cdot MO \) after simplifying through proportional triangles.

### c) Proving that \( I \) is the midpoint of \( BK \).

Since \( MC \) intersects \( BK \) at \( I \), and we already established that \( BK \) is perpendicular to \( AC \), we can create a relationship using congruent triangles or by symmetry of the circles involved.

- \( BK \) divides it equally at \( I \) because both segments \( BI \) and \( IK \) share equal angles, ensuring that \( I \) is the midpoint.

Thus, the points \( M, A, O, B \) are concyclic, the equation holds, and \( I \) is confirmed as the midpoint of \( BK \).