simple looking but hard to prove geometrical problem: prove that 4 points on the same circle.

First, let's prove an intermediate conclusion, or a lemma, which can be stated as follows.

Lemma Let $l$ be another exterior common tangent (namely, not $CD$) of the circles $(ADE)$ and $(BCE)$. Then $l$ is tangent to the circle $(ABE)$.

Proof All points are labeled as the figure shows. Notice that \begin{align*} AK&=AO-KO=\frac{1}{2}(AD+AE-DE)-KJ,\\ BL&=BN-LN=\frac{1}{2}(BC+BE-CE)-LM.\\ \end{align*} Hence \begin{align*} AK+BL&=AB+\frac{1}{2}(AE+BE-DE-CE)-(KJ+LM)\\ &=AB+\frac{1}{2}(AE+BE-CD)-(JM-KL)\\ &=AB+\frac{1}{2}(AE+BE-CD)-(PQ-KL)\\ &=AB+KL+\frac{1}{2}(AE+BE-CD)-(EP+EQ)\\ &=AB+KL,\\ \end{align*} which shows that the quadrilateral $ABLK$ has an inscribed circle. Apparently, it must be the one of triangle $AEB$, namely, $ABLK$ and $ABE$ have the identical inscribed circle. The conclusion we want to prove is followed. Moreover, we may see that, $AE,GH,l$ and $BE,GF,l$ are respectively concurrent.

Now, come back to deal with the present problem. Notice that $GA||EH$ and $GB||EF$. hence $\angle HEF=\angle AGB$. But $ABLK$ is a circumscribed quadrilateral, then it's obvious that $\angle AGB+\angle KGL=180^o$. As a result, $\angle HEF+HGF=180^o$, which implies that $H,E,F,G$ are concyclic. We are done.