2. Because m is the midpoint of arc AB, equal arc and equal angle, ∠ BCM = ∠ MAB = 45.

3. Angle COB = 2, angle OCB+ angle PCB = 90, and △BCO is equilateral △, so we can get angle COB = 2 and angle PCB = 60, so ∠ CAB = 30.

4. In triangle NBC, AB=4, then bc=2. ∠CBN=60, ∠BCN=45, the rest are based on cosine formula, but this triangle has been determined, so we can find BN and an.

MN times MC=BN times AN.

Watch it again by yourself.

I remember there is a theorem that determines that triangle ABC is a right triangle.

As long as the triangle ABC is determined as a right triangle

You can find the second small problem.

The base (CB) of the triangle is equal to the hypotenuse (AB) of 1/2.

From the second step, the triangle CBO can be determined to be an equilateral triangle.

This length can determine that the CBP of the foot is 120.

Since AC=PC, the triangular cap is an equilateral triangle, and AO=OB=BP=CB.

Although it can be determined that the triangle CBP is an isosceles triangle with an obtuse angle of 120, the angle BCP= the angle BPC=30.

Since the angle PCB = 30, the angle BCO = 60, and the angle PCO= angle PCB+ angle BCO = 60°+30° = 90°.

So PC is the tangent of circle O.