@The-Darkin-Blade Hi there!
Thank you for mentioning about this voice typo! We will definitely take a look and think about what to do about it!
Now, let me answer your question about inscribed angles.
The formal wording of this theorem goes like this:
Inscribed Angle Theorem:
Any angle that is inscribed in a circle is half of the central angle that subtends the same arc on the circle.
Proof:
Let our inscribed angle be ∠BAC and the central angle ∠BOC.
There are two cases for the position of the inscribed angle relative to the central angle.
Case 1:

Let's draw one more radius, OA. Now we have three isosceles triangles: △AOB, △AOC and △BOC. So the the angles at the base are equal. Let α=∠BAO=∠ABO, β=∠CAO=∠ACO and γ=∠BCO=∠CBO. So the ∠BAC=α+β.

Let's now find the measure of the angle ∠BOC:
The sum of the angles in any triangle is equal to 180∘, so ∠BOC=180∘−2γ and also ∠BAC+∠ABC+∠ACB=180∘.
On the other hand, sum of the angles in △ABC=2α+2β+2γ=180∘, hence 2α+2β=180∘−2γ=∠BOC. And we got what we wanted: ∠BOC=2α+2β=2(α+β)=2∠BAC.
So we are done!
Case 2:
Now, if you want, you can try to prove this case yourself, using the ideas from the proof of the first case 

Let's draw one more radius, OA. Now we have three isosceles triangles: △AOB, △AOC and △BOC. So the angles at the base are equal. Let α=∠BAO=∠ABO, β=∠CAO=∠ACO and γ=∠BCO=∠CBO. So the ∠BAC=α−β.

Let's now find the measure of the angle ∠BOC:
The sum of the angles in any triangle is equal to 180∘, so ∠BOC=180∘−2γ and also ∠BAC+∠ABC+∠ACB=180∘.
On the other hand, sum of the angles in △ABC=(α−β)+(α+γ)+(γ−β)=2α−2β+2γ=180∘, hence 2α−2β=180∘−2γ=∠BOC. And we got what we wanted: ∠BOC=2α−2β=2(α−β)=2∠BAC.
So we are done again, using the same algorithm!