Let $K$, $L$ be the feet of the perpendicular lines drawn from $B$ to the side $CA$ and from $C$ to the side $AB$, respectively. Since the line segment $AB$ is a diameter of the circumcircle to the triangle $ABD$, $\angle AEB = 90^\circ$. We see that the triangles $AEB$ and $ALE$ are similar, since they have the angles of same magnitudes. Therefore, we have $AE : AL = AB : AE$, and we get $AE^2 = AB \cdot AL$.

Similarly, we obtain $AF^2 = AC \cdot AK$ from the similarity of the triangles $AFC$ and $AKF$. Since $\angle BKC = \angle BLC = 90^\circ$, we see that the points $B$, $C$, $K$, $L$ lie on the circumference of a same circle. Using the well-known theorem on the power of a point with respect to a circle, we then obtain $AB \cdot AL = AC \cdot AK$, and therefore, we obtain $AE^2 = AF^2$, which shows that $AE = AF$.