Question - Tangent and Secant Angle Relationships in a Circle

Consider triangle AEB and AED:

\angle AEB = \angle AED = 90^{\circ} \text{ (angle formed by a tangent and a chord is a right angle)}

Consider triangle ABE and the exterior angle \angle EAX:

\angle EAB + \angle ABE = \angle EAX

\angle ABE = \frac{\angle BCD}{2} \text{ (angle at the center is twice the angle at the circumference)}

Now consider the quadrilateral BCED:

\angle BCD + \angle BED = 180^{\circ} \text{ (opposite angles of a cyclic quadrilateral sum to 180 degrees)}

\angle BED = 180^{\circ} - \angle BCD

Solve for \angle BCD using the fact that \angle AED = 90^{\circ}:

\angle AED = \angle BED = 180^{\circ} - \angle BCD

90^{\circ} = 180^{\circ} - \angle BCD

\angle BCD = 90^{\circ}

So, \angle ABE = \frac{\angle BCD}{2} = \frac{90^{\circ}}{2} = 45^{\circ}

Now, solve for \angle EAX:

\angle EAX = \angle EAB + \angle ABE

\angle EAX = 80^{\circ} + 45^{\circ}

\angle EAX = 125^{\circ}

Consider the exterior angle \angle AXF for triangle AXF:

\angle AFX + \angle FAX = \angle EAX

\angle AFX = \angle EAX - \angle FAX

\angle AFX = 125^{\circ} - 80^{\circ}

\angle AFX = 45^{\circ}