I'll do problem 9 to get you started.
Refer to figure 1 shown below. I've highlighted triangle AOB in red. This is an isosceles triangle with AO = OB as the congruent sides. These sides are congruent because they are radii of the same circle. Consequently, it means that the base angles opposite those sides are congruent.
I'll let x be the base angles. I'll let the remaining interior angle be y. Furthermore, z is adjacent to angle y.
For triangle AOB, and any triangle really, the interior angles add to 180 degrees.
A+O+B = 180
x+y+x = 180
2x+y = 180
y = 180-2x
Then we can say
(angle AOB) + (angle AOP) = 180
y + z = 180
180-2x + z = 180
z = 180+2x - 180
z = 2x
The punchline here is that angle ABP doubles to angle AOP.
angle AOP = 2*(angle ABP)
You'll repeat these same type of steps for triangle BOC in figure 2. The steps are effectively identical so there's not much for me to explain that I already haven't done so. You should find that r = 2m which shows angle PBC doubles to angle POC.
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To wrap everything up for problem 9, we will combine the results we found for variables z and r.
angle AOC = z+r
angle AOC = 2x + 2m
angle AOC = 2(x+m)
angle AOC = 2*(angle ABC)
This concludes the inscribed angle theorem proof. Any inscribed angle doubles to get the corresponding central angle, where both subtend the same arc. In this case, they both subtend the arc APC.