Prove that $\angle DIH=30^{\circ}$ Let $ABCD$ a parallelogram with $\angle A>60^{\circ}$. If $\triangle ACE, \triangle DEF, \triangle BEG$ are equilateral as below then show that $\triangle DHI$ is isoscele, where $FB\cap DG=\{H\}, FB\cap DC=\{I\}$.
I notice that $\triangle DEG\equiv \triangle BFE$. Then $\angle HDE\equiv \angle HFE$ and $\angle HBE\equiv \angle HGE$. So $HDFE, HBGE$ are inscriptible. Then $\angle DHI=120^{\circ}$. Now I have to prove that $\angle DIH=30^{\circ}$.

 A: Observe that $\triangle ABE$ and $\triangle CEG$ are congruent. Hence $DC=AB=CG$. Let $\angle ACD= x$. Then $\angle EAB= \angle ECG= 60 ^ {\circ} +x$. On the other hand we have $\angle DCE=  60^{\circ} -x$. Therefore $\angle DCG = 120 ^{\circ}$. Since $DC=CG$, we are done.
A: Kill a fly with a bazooka.
Suppose $C$ and $D$ are fixed and we move $A$ and $B$.
Let $T$ be translation for a vector $\overrightarrow{CD}$, let $R_C$ be a rotation around $C$ for $60^{\circ}$ and $R_D$ be a rotation around $D$ for $-60^{\circ}$.
Fact 1: Compositon $I:= R_C\circ T$ takes $B$ to $E$. Since $I$ is a compostion of rotation and translation $I$ must also be a rotation around fixed point for angle $60^{\circ}$ and that must be $G$. So $G$ must be fixed.
Fact 2: Composition $J: = R_D\circ R_C\circ T$ takes $B$ to $F$ and since $R_D\circ R_C$ is rotation for an zero angle it must be a translation, so $J$ is composition of two translation so it must be translation, which means $\overrightarrow{BF}$ is fixed vector.
Now the problem must be easy to finish. Just take some nice configuration and you are done.
A: (This is the same as the deleted question 4594184... There i was staring an answer, but it was removed while trying to give a picture. A lot of work is gone. Here is a retyped version... In fact, the two pictures below are all what is needed to conclude.)
Construct $J$ as in the following picture so that $\Delta DBJ$ is equilateral, $J$ and $E$ being in the opposite halpf-planes w.r.t. $DB$, so that $\color{blue} H$ becomes the Fermat point of $\color{blue}{\Delta DBE}$:

As known, inscriptible quadrilaterals are built, two of them are mentioned in the question, and the three lines through the Fermat point $H$ ($DHG$, $EHJ$,
$FHB$) divide the full $360^\circ$-angle around $H$ in six $60^\circ$-angles. In particular, $HJ$ is the angle bisector in $\Delta DHI$, so it is enough to show $EHJ\perp DIC$ to conclude. For this last step a lot of points are irrelevant, so isolating the needed ones we have to show in the lighter picture

involving the parallelogram $ABCD$ with diagonals intersecting in $O$, and the points $E,J$ that $EJ$ is perpendicular to the direction $CD$.
To see this consider the $90^\circ$-rotation $r$ centered in $O$ that maps rays as follows, $r(OC)=OE$, $r(OD)=OJ$, and the scaling $s$ centered in $O$ with factor $\sqrt 3=OE:OC=OJ:OD$. Then the composition $rs=sr$ maps $\Delta OCD$ to $\Delta OEJ$, thus extracting the information form the rotational part
$$  
 CD\perp EJ\ .
$$
$\square$
