Estonian Math Competitions

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.

---

Alternative solution.

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