Berkeley Math Circle: Monthly Contest 8

Solution:

Since $AE$ and $AF$ are tangents from the same point to the same circle, we must have $AE = AF$. Thus, $\triangle AEF$ is isosceles, so
$$\angle AEF = \angle AFE = \frac{180^\circ - \angle A}{2} = 90^\circ - \frac{1}{2} \angle A.$$
Now, since $AE$ is a tangent to the circle, $\angle EDF = \angle AEF = 90^\circ - \frac{1}{2} \angle A$. Thus, $\angle EDF < 90^\circ$. Similarly, the other two angles must also be less than $90^\circ$, so the triangle is acute.