$D$ is the A-excenter of $\triangle BAC$. Let $I$ be the incenter of $\triangle BAC$ or simply center of $\Omega$. Now we know that $\angle IBD=\angle ICD=90$. As $\angle ICD+\angle IBD=180$ ,$ I,C,B,D$ concylic. Now as $DE$ tangent to $\Omega$ so $\angle IED=90$. As $\angle IED=90=\angle IBD$ so $I,E,B,D$ concylic. Thus $I,E,B,D,C$ concyclic. Now $\angle BEC=\angle DEB+\angle DEC=\angle BCD+\angle DBC= 90-\angle IBC+90-\angle ICB = 180- ( \frac{\angle ABC}{2}+\frac{\angle ACB}{2})=180-(\frac{\angle ABC+\angle ACB}{2})=180-(\frac{180-\angle BAC}{2})= 180-(\frac{140}{2})=\fbox{110}$.