, 2008

Let $T$ be the intersection of the chords $DE$ and $AB$. We know that $DTB$ is a right triangle. First, assume that the triangle $ABE$ is isosceles and write $\angle AEB = \angle BAE = \alpha$. Inscribed angles $\angle EAB$ and $\angle EDB$ over $BE$ are equal, so $\angle EDB = \alpha$. Since $DE$ is perpendicular to $AB$, we have $\angle ABD = \frac{\pi}{2} - \alpha$. In the right triangle $ABC$ we have $\angle ABC = \frac{\pi}{2} - \alpha$, so $\angle CAB = \alpha$ and the angle $\angle CAB$ between the line $AC$ and the segment $AB$ is equal to the angle $\angle AEB$ over the chord $AB$. Thus, $CA$ is tangent to the circumcircle $K$.

Conversely, assume that $AC$ is tangent to $K$. Let $\angle CAB = \alpha$. The angle $\angle BAC$ is equal to the inscribed angle $\angle AEB$ over the chord $AB$, so $\angle AEB = \alpha$. Also, $\angle ABC = \frac{\pi}{2} - \alpha$, so $\angle TDB = \alpha$. We see that $\angle EDB = \alpha$ and this angle is in turn equal to $\angle BAE$, because they are both inscribed angles over the same chord $BE$. We have $\angle BAE = \alpha = \angle BEA$ and the triangle $ABE$ is isosceles with the apex at $B$.