Hi @vibrantrobin !
When you use \(\frac{a-b}{2}\), remember that \(a\) is the arc opposite of the angle on the other side of the circle, while \(b\) is the arc close to the circle.
d9bae2ff-b909-4064-b767-135d91ab14f5-image.png

Here's a visual of what the formula looks like:
e72567fb-85c3-4b3e-beb2-cc0edf4fdfc9-image.png

Do you see how, in the original problem, the larger arc is \(\frown \atop AB \) \( = 130^{\circ} \) while the smaller arc is the \(30^{\circ}\) arc. Then we can just do \(\frac{130^{\circ}-30^{\circ}}{2} = 50^{\circ}\) 🙂

Great question, @vibrantrobin !
So, the solution uses two theorems. For \(\angle FEH\), it uses the Inscribed Angle Theorem.
a28fa1dc-e976-4817-b60a-20f51723ed54-image.png
As long as the vertex of the angle (in this diagram, it's point \(B\)) is on the circle, the degree measure of the angle is equal to half of the intercepted arc. Using this, in your problem, \(\angle FEH=\frac12 \widehat{FGH}\) (this isn't the correct symbol for arc but it's all I could find).

For \(\angle BFI\), it uses another theorem. It's a special case of when a tangent and a secant interesect ON the circle.
ba66b038-10cc-4dca-8902-79f193373ba9-image.png
Whatever arc the two lines cut out, the angle is half of the arc measure. Using this, in your problem, \(\angle BFH = \angle BFI = \frac12 \widehat{FGH} \).

Since they are both equal to \(\frac12 \widehat{FGH}\), they are equal!

There are a lot of theorems that involve secants and tangents, like these three for example:
9c414e21-f113-4c 7a-bb14-68feee82984e-image.png
You should look more into it if you're interested! Hope this helps 🙂