Here's my solution: Claim: We claim that $\boxed{\angle DHE = \dfrac{5\pi}{9}}$. Proof: Firstly since $ADEC$ is cyclic $\implies \angle BCA = \pi - \angle EDA = \angle BDE$. Similarly, $\angle BED = \angle BAC$. Hence $\triangle BDE$ is similar to $\triangle BCA$. For brevity, we define $\angle BAC =\alpha $, $\angle BCA =\gamma$ and $\angle HDE = \angle HED = \theta$. Lemma: $\theta = \dfrac{2\pi}{9}$. Proof: We apply the law of sine in $\triangle BDH$ and $\triangle BEH$. $\longrightarrow \dfrac{BH}{sin(\gamma + \theta)} = \dfrac{DH}{sin(\dfrac{\pi}{2} - \alpha)}$......(1). $\longrightarrow \dfrac{BH}{sin(\alpha + \theta)} = \dfrac{EH}{sin(\dfrac{\pi}{2}-\gamma)}$......(2). Combining both the equations yields: $\dfrac{sin(\alpha + \theta)}{sin(\dfrac{\pi}{2} - \gamma)} = \dfrac{sin(\gamma+\theta)}{sin(\dfrac{\pi}{2} - \alpha)} \iff \dfrac{sin(\alpha + \theta )}{sin(\alpha + \dfrac{2\pi}{9})}= \dfrac{sin(\gamma + \theta)}{sin(\gamma + \dfrac{2\pi}{9})}$. Now as $\alpha \neq \gamma$ and $\theta$ is acute $\implies \theta = \dfrac{2\pi}{9}$. And hence the claim. P.S. One can also claim that $H$ is the circumcenter of $\triangle BDE$ and proceed in that direction.

let $O$ be the circumcenter of $BDE$ and $K $ is the foot of the $O$ on $BA$ and $F=BO \cap AC$ then $\angle BOK=\angle BED =\angle BAC$ i.e $OKFA$ is cyclic thus $BF\perp AC\implies H \in BO $ but $DH=HE $ and BDE scalene hence $H=O\implies \angle DHE =2( 180- \angle DBE)=100$