parmenides51 wrote: Outside the parallelogram $ABCD$ on its sides $AB$ and $BC$ are constructed right triangles $ABK$, and $BCM$. Prove that the triangle $KMD$ is equilateral. ????

given triangles are equilateral and not right ones, translation error, thanks for noticing

Look at https://www.youtube.com/watch?v=WuF6F9Ki8-E

We have $\angle{KBM} = 240^{\circ} - \angle{ABC} = 60^{\circ} + \angle{DAB} = 60^{\circ} + \angle{BCD}$ and $AK = KB = CD, AD = MC = MB$. Then $\triangle KAD = \triangle DCM = \triangle KBM$ or $KD = DM = MK$