Proof in Circle geometry question

Question

Let Γ be a semicircle with diameter . The point lies on the diameter and points and lie on the arc , with between and . Let the tangents to Γ at and meet at . Suppose that ∠ = ∠. Prove that ∠ = ∠ + ∠.

I got this question from a sample competition paper aimed at 16-18 year olds. I think that I can prove this by proving that the point C must be at the circle's centre by contradiction; if it wasn't then the 2 angles which are supposedly equal wouldn't in fact be equal.

I have 3 questions:

1. Is my argument correct?

2. If my argument is correct, what would be a rigorous proof for my argument?

3. Is there an alternative method? If so please hint to it.

Thank you for your help.

Answer 1

It is certainly not the case that $C$ must be the center. Consider the diagram below:

Here $C$ is not the center. However $\angle EFD = \angle ACD + \angle ECB$ still holds.

To show that equality, it suffices to show that $CEFD$ is cyclic.

This can be accomplished by proving the orange angles below are equal, and that is my hint. (There could be alternative methods but I'm not sure)

We will need to show that $CE = CG$ given that $\angle ECB = \angle GCB$.

Consider the center of the circle, $O$. $OE = OG$ as radii, with $OC$ as common sides, leads to $\triangle COE \cong \triangle COG$ by "long side-short side-angle".

We can also see this by dropping perpendiculars from $O$. Then $\triangle COP \cong \triangle COP'$ by AAS, and $\triangle POE \cong \triangle P'OE$ by RHS. Either way, we have $CE = EG$, and we can proceed.