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 $\angle DIH=30^{\circ}$.
Best Answer
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.