Question - Geometry Problem Involving Tangents and Circles

Let $\angle ABC = \angle BAD = y$ since they are angles subtended by the same arc AD. Similarly, let $\angle ACB = \angle ADB = z$ since they are subtended by the same arc AB.

Since AE and AX are tangents to the circle at points B and C, and tangents from a point outside a circle are equal, we have AE = AB and AX = AC.

Triangle ABE is isosceles with AE = AB, so $\angle ABE = \angle AEB = y$. Similarly, triangle ACX is isosceles with AC = AX, so $\angle ACX = \angle AX = z$.

Angles around point A add up to 360°, thus $\angle EAX + 2y + 2z = 360°$.

Substitute $\angle EAX = 80°$ to find $2y + 2z = 280°$.

Divide by 2 to find $y + z = 140°$.

The angle sum in triangle ABC is 180°, thus $y + z + \angle BAC = 180°$.

Substitute $y + z = 140°$ into the previous equation to find $\angle BAC = 40°$.

Now, focusing on quadrilateral ADBC, notice that $\angle ABC + \angle BAD + \angle BAC + \angle BCD = 360°$.

Since $\angle BAC = 40°$, $\angle ABC + \angle BAD = 2y = 140°$, and $\angle BCD = 90°$ (angle in a semicircle), we substitute these into the quadrilateral angle sum to find $2y + 40° + 90° = 360°$.

Combine like terms to get $2y + 130° = 360°$.

Subtract 130° from both sides to solve for $2y$, yielding $2y = 230°$.

Divide by 2 to find $y = 115°$.

Finally, $\angle AFX = 115°$ because it is equal to $\angle ABC$ which is equal to $y$.