if CE is a median note CH<=CE. suppose angel B<angle C : then BD>CE>=CH, contradiction so, angle B=C and equilateral golry to god tony BARKER

Connect $ED$. Because $DA=DC$ and $\angle AED = 90^\circ$, $\triangle ADE$ is equilateral. And $EC=DB$, $\angle BED = 120^\circ = \angle EDC$, so $\triangle EDB $ ≌ $\triangle DEC$. Then $EB=DC$, $AB=AC$, and $\angle BAC = 60^\circ$. So $\triangle ABC$ is equilateral.