The conditions of the problem imply that $DE$ is the midsegment parallel to the side $BC$ of the triangle $ABC$ (Fig. 30).

By properties of inscribed angle, the line $AB$ is tangent to the circumcircle of the triangle $BEC$ if and only if $\angle ECB = \angle DBE$, and the line $AC$ is tangent to the circumcircle of the triangle $BED$ if and only if $\angle DBE = \angle AED$. But $\angle ECB = \angle AED$ since the lines $DE$ and $BC$ are parallel, whence validity of either of these two equalities implies validity of the other one.

By powers, the line $AB$ is tangent to the circumcircle of the triangle $BEC$ if and only if $AB^2 = AE \cdot AC$, and the line $AC$ is tangent to the circumcircle of the triangle $BED$ if and only if $AE^2 = AD \cdot AB$. As $AB = 2AD$ and $AC = 2AE$, the equality $AB^2 = AE \cdot AC$ is equivalent to the equality $2|AD|^2 = |AE|^2$, as well as the equality $AE^2 = AD \cdot AB$ is equivalent to the equality $AE^2 = 2AD^2$. Hence both conditions reduce to the same equality, which solves the problem.