Segment AC is the diameter of the circle, AB is the chord, MA is the tangent, the angle MAB

Segment AC is the diameter of the circle, AB is the chord, MA is the tangent, the angle MAB is acute. Prove that angle MAB = angle ACB.

Since AM is tangent to the circle, and OA is the radius drawn to the point of tangency, then AM is perpendicular to OA.

The ABC triangle is rectangular, since its hypotenuse AC coincides with the diameter of the circle.

Let the angle BAC = α, then the angle ACB = (90 – α).

Since the angle MAO = 900, then the angle MAW = (90 – BAC) = (90 – α).

Then the angle МАВ = АСВ = (90 – α), as required to prove.