vote I'm not sure which theorem of Thales you are thinking of, but since you know that $\overline{DE}$ is parallel to $\overline{BC}$, $\overline{AB}$ is a transversal , so we have that $\angle ADE=\angle ABC$. Obviously, $\angle DAE=\angle BAC$, therefore, $\triangle ABC$ is similar to $\triangle ADE$. Since $\overline{AD}$ is half the